PH 105 Exam 1 HW

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The position of a softball tossed vertically upward is described by the equation y=c₁t-c₂t² where y is in meters, t in seconds, c₁= 3.39 m/s, and c₂= 3.04 m/s². Find its velocity at t=1.03 s.

-2.8724 m/s explanation: Let: t=1.03 s v= 3.39 m/s - 2(3.04 m/s²) (1.03 s)= -2.8724 m/s

A place kicker must kick a football from a point 32 m (about 35 yd) from the goal, and the ball must clear the crossbar, which is 3.05 m high. When kicked, the ball leaves the ground with a speed of 24 m/s at an angle of 55° to the horizontal. To determine if the ball approaches the cross-bar while still rising or while falling, what is its vertical velocity at the crossbar?

-3.12139 m/s explanation The vertical component of the velocity is vᵧf = v₀sinθ- g∆t = v₀sinθ- g ((∆x)/v₀cosθ)) = (24 m/s)sin55° - ((9.8 m/s²)(32 m))/((24 m/s) *cos55°) = -3.12139 m/s

A book weighing 17 N is placed on a table. What is the net force on the book in this case?

0 N explanation The book is stationary, so the net force is 0

The horizontal surface on which the block of mass 3.1 kg slides is frictionless. The force of 30 N acts on the block in a horizontal direction and the force of 60 N acts on the block at an angle as shown below. LOOK AT FIGURE 14 IN GOOGLE DOC What is the magnitude of the resulting acceleration of the block? The acceleration of gravity is 9.8 m/s².

0 m/s² Let: F₁= 30 N F₂= 60 N α= 60° m= 3.1 kg and g= 9.8 m/s² The force F₂ has a horizontal component F₂cosα acting to the left, and the force F₁ acts to the right, so Fnet= ma= F₂cosα-F₁ =F₂cos60°-F₁ a= ((F₂)(½)-F₁)/m = (60 N *½ -30 N)/3.1 kg= 0

A(n) 85 kg block is pushed along the ceiling with a constant applied force of 1100 N that acts at an angle of 70° with the horizontal, as in the figure. The block accelerates to the right at 3.8 m/s² LOOK AT FIGURE 22 ON GOOGLE DOC The acceleration of gravity is 9.8 m/s². What is the coefficient of kinetic friction µ between the block and the ceiling?

0.26523 explanation Solving Eq. 2 for µ, we have µ= (1/N) *(Fcosθ-ma) Substitute (3) into (4) µ= (Fcosθ-ma)/(Fsinθ-mg) =((1100 N)cos70° - (85 kg)(3.8 m/s²))/((1100 N)sin70°- (85 kg)(9.8 m/s²)) = 0.26523 To check this result, let us solve for F F= ((a-µg)/(cosθ-µsinθ))*m =((3.8 m/s²)-(0.26523)(9.8 m/s²))/((cos70°-(0.26523)sin70°)) *85 kg = 1100 N

A block is released from rest on an inclined plane and moves 2.5 m during the next 4 s. The acceleration of gravity is 9.8 m/s². LOOK AT FIGURE 19 ON GOOGLE DOC What is the coefficient of friction µₖ for the incline?

0.310603 explanation Applying Newton's Second Law of Motion ∑F₀= ma and Eq.1 and the sine component of the weight acts down the plane and friction acts up the plane. The block slides down the plane, so ma=mg*sinθ- µₖ*mgcosθ (2l)/(t²)= g(sinθ-µₖ*cosθ) 2l= gt² (sinθ-µₖ*cosθ) µₖ= (gt²sinθ-2l)/(gt²cosθ) = tanθ - (2l)/(gt²cosθ) = tan19° - ((2*2.5 m)/((9.8 m/s²)(4 s)²*cos19)) =0.310603

A block is released from rest on an inclined plane and moves 2.5 m during the next 4 s. The acceleration of gravity is 9.8 m/s². LOOK AT FIGURE 19 ON GOOGLE DOC What is the magnitude of the acceleration of the block?

0.3125 m/s² explanation m= 11 kg l= 2.5 m θ = 19° and t= 4 s Draw a FBD for the block The acceleration can be obtained through kinematics. Since v₀=0 l=v₀t + ½at²= ½at² a= (2l)/t² ((2(2.5 m))/(4 s)² = 0.3125 m/s²

A 4 kg block rests on a horizontal table, attached to a 2 kg block by a light string as shown in the figure. The acceleration of gravity is 9.81 m/s². LOOK AT FIGURE 21 ON GOOGLE DOC What is the minimum coefficient of static friction such that the objects remain at rest?

0.5 explanation Let: m₁= 4 kg and m₂= 2 kg In the absence of friction, the 4 kg box will move to the right, and the 2 kg box will move down, so the friction force f acts to the left. Consider the forces acting on each object from a FBD. The objects remain at rest. The mass on the table is in equilibrium vertically: ∑Fᵧ= m₁a₁ᵧ= 0 N-m₁g= 0 N=m₁g and horizontally: ∑Fₓ= m₁a₁ₓ= 0 T-f=0 T=f=µₛN= µₛm₁g The hanging mass is in equilibrium vertically: ∑Fᵧ= m₂a₂ᵧ= 0 T-m₂g=0 T=m₂g Thus T=µₛm₁g = m₂g µₛ= m₂/m₁= 2 kg/4 kg = 0.5

You are on a hike in the mountains. You have 3.06 km left to go before your next campsite. The sun will set in 5.27 h. What average speed must you walk to get to camp at sunset?

0.580645 km/h explanation: d= 3.06 km and t= 5.27 h d= rt or r=(d/t) = (3.06 km/5.27 h)= 0.580645 km/h

A box of books weighing 345 N moves with a constant velocity across the floor when it is pushed with a force of 425 N exerted downward at an angle of 35.1° below the horizontal. Find µₖ between the box and the floor.

0.589968 Fₓ= Fcosθ Fᵧ= Fsinθ Fₖ= µₖFn Parallel to the floor: Fₓnet= Fappliedₓ- Fₖ= 0 Perpendicular to the floor: Fᵧnet= Fn-Fappliedᵧ- F(g)= 0 Given: F(g)= 345 N Fapplied= 425 N θ= 35.1° Consider the forces parallel to the floor: Fₖ=Fappliedₓ =Fappliedcosθ = (425 N)cos35.1° =347.714 N Consider the forces perpendicular to the floor: Fn = F(g) +Fappliedᵧ = F(g) + Fappliedsinθ = 345 N + (425 N) sin35.1° = 589.377 N Thus the coefficient of friction is µₖ= (Fₖ)/(Fn) = (347.714 N)/(589.377 N) = 0.589968

A ball is thrown upward from the ground with an initial speed of 40 m/s; at the same instant, a ball is dropped from a building 26 m high. After how long will the balls be at the same height? The acceleration of gravity is 9.8 m/s².

0.65 s explanation: Let: v(up)= 40 m/s h= 26 and g=-9.8 m/s² y(down)= h+ 1/2 gt² and y(up) = v(up)*t + 1/2 gt² The two objects pass one another when y(down)=y(up) h-1/2 gt²= v(up)t-1/2 gt² h=v(up)*t t= h/v(up)= (26 m)/(40 m/s) = 0.65 s

A ball is thrown upward. LOOK AT FIGURE 5 ON GOOGLE DOC What is the time interval between the release of the ball and the time it reaches its maximum height? The acceleration of gravity is 9.8 m/s² and the maximum height is 4.7 m. Neglect air resistance.

0.979379 s explanation: Consider the motion upward. Let: vf= 0 m/s Δh = 4.7 m and g= -9.8 m/s² vf= v₀ + gt = 0 v₀ = -gt and Δy=v₀t + 1/2 gt² = -gt² + 1/2 gt²= -1/2gt² t(up)= √(-2Δy/g)= √(-2(4.7 m)/-9.8 m/s²) = 0.979379 s

A gram is what part of a kilogram? 1) 0.001 2) 0.01 3) 0.1 4) 0.5

1) 0.001 - conversions!

Consider the following graph of motion. LOOK AT FIGURE 3 ON GOOGLE DOC How many meters can Swimmer 1 cover in 30 seconds? 1) 60 m 2) 10 m 3) 80 m 4) 50 m 5) 70 m 6) 40 m 7) 90 m 8) 20 m 9) 30 m 10) 100 m

1) 60 m explanation: In the first 30 s, Swimmer 1 has moved from 0 m to 60 m

A block starts at the bottom of a ramp making an angle of θ with the horizontal, at initial speed v₀. As it slides up it slows due to gravity and kinetic friction, eventually coming to a stop. Static friction is not large enough to hold it at rest and it slides back down the ramp. LOOK AT FIGURE 16 ON GOOGLE DOCS When it reaches its initial position, compare its speed to the initial speed v₀. 1) Less than v₀ 2) Without knowing µₛ and µₖ we cannot state precisely how the final speed compares to v₀ 3) Even greater than v₀ because of gravity 4) Not enough information has been provided 5) The same as v₀

1) Less than v₀ explanation When the block is sliding up the plane, the force of kinetic friction and the component of the force of gravity parallel to the plane are in the same direction, so the deceleration of the block as it travels upward along the plane is large. But when the block is sliding down the plane, the force of kinetic friction is in the opposite direction to the component of the force of gravity parallel to the plane, so the net force along the plane is much less than in the case where it moves upward. As a result, since it starts from rest in sliding down the plane, it is moving slower when it reaches the bottom as compared to its initial speed starting upward.

A particle starts from rest, initially at x(0)=0, and moves with velocity vₓ(t)=Bt⁴ What is the particle's acceleration? 1) aₓ=4Bt³ 2) aₓ=B by definition 3) The acceleration is the slope of vₓ(t), but since this is not a constant, the acceleration is undefined 4) aₓ=(B/5) t⁵ 5) aₓ= (vₓ²/2x)= (B²t⁸)/(2x) when the particle is at the position x 6) None of these

1) aₓ=4Bt³ explanation: aₓ= (dvₓ)/(dt) = (d/dt)(Bt⁴)= 4Bt³

Consider a person standing in an elevator that is accelerating upward. The upward normal force N exerted by the elevator floor on the persons is 1) larger than the weight (gravitational force) of the person 2) smaller than the weight (gravitational force) of the person 3) identical to the weight (gravitational force) of the person

1) larger than the weight (gravitational force) of the person explanation According to Newton's 2nd Law, the equation of motion for the person on the elevator is N-W= ma, so N= W + ma > W

How would your mass change if you took a trip to the space station? 1) no change in mass 2) increases; you weigh more 3) decreases; you weigh less

1) no change in mass explanation Your mass never changes. You would weigh slightly less because you are farther from the center of the Earth

Consider the instantaneous velocity of a body. This velocity is always in the direction of 1) the motion at that instant 2) the least resistance at that instant 3) the net force at that instant

1) the motion at that instant explanation: The instantaneous velocity vector will always point in the direction of the motion at that instant

You have two forces, 98.3 N and 71.8 N. What is the direction of the resultant factor? 1) upward 2) downward

1) upward explanation The direction of the resultant will be that of the larger force

The microliter is a unit of 1) volume 2) length 3) mass 4) temperature

1) volume - the microliter is a metric unit of a volume

A machine in an ice factory is capable of exerting 2.96 x 10² N of force to pull large blocks of ice up a slope. The blocks each weigh 1.29 x 10⁴ N. Assuming there is no friction, what is the maximum angle that the slope can make with the horizontal if the machine is to be able to complete the task?

1.31481° explanation Fnet= Fapplied - F₉‖= 0 F₉‖= F₉sinθ Given: Fapplied= 2.96 x 10 ² N F₉= 1.29 x 10⁴ N sinθ = Fapplied/F₉ θ= sin^-¹ (Fapplied/F₉) = sin^-¹ (296 N/12900 N) = 1.31481°

Predict the distance Swimmer 2 can go in 60 seconds. LOOK AT FIGURE 3 ON GOOGLE DOC 1) 80 m 2) 70 m 3) 110 m 4) 130 m 5) 50 m 6) 120 m 7) 40 m 8) 100 m 9) 90 m 10) 60 m

10) 60 m explanation: There is an increase of 10 m every 10 s, so the distance would be 60 m

Find the dimensions for the quantity c in the expression: θ=s/c 1) [c]=[L][T] 2) [c]=1/[T] 3) [c]=[L][T²] 4) [c]=[T]/[L] 5) [c]=[L]/[T²] 6) [c]=[T] 7) [c]=1/[L] 8) [c]=[L²] 9) [c]=[L]/[T] 10) [c]=[L]

10) [c]=[L] explanation: c=s/θ [c]=[s]/[θ]= [s]/rad= [L]

It is given that r and s are distances with unit [L], t is a time with unit [T] and θ is an angle in radians. Find the dimensions for the quantity c in the expression: s=ct 1) [c]=[L²] 2) [c]=[T]/[L] 3) [c]=1/[L] 4) [c]=[T] 5) [c]=1/[T] 6) [c]=[L]/[T²] 7) [c]=[L][T] 8) [c]=[L][T²] 9) [c]=[L] 10) [c]=[L]/[T]

10) [c]=[L]/[T] explanation: c = (s/t) [c]=[s]/[t]= [L]/[T]

Consider a force F= 634 N pulling 3 blocks of masses m₁= 9.66 kg, m₂= 15.7 kg, and m₃= 38 kg along a frictionless horizontal surface. LOOK AT FIGURE 15 IN GOOGLE DOC Find the acceleration of the blocks

10.0063 m/s² explanation The three blocks move together as a single body, so as far as the Newton's Second Law is concerned, we may treat them as a single body of net mass Mnet= m₁ + m₂ + m₃= 63.36 kh Hence, Mneta= F(ext)= F= 634 N regardless of the internal froces T₁ and T₂, and therefore a= F/Mnet= (634 N)/(63.36 kg)= 10.0063 m/s²

A Knight of the Round Table fires off a vat of burning pitch from his catapult at 13.3 m/s, at 39⁰ above the horizontal. The acceleration of gravity is 9.8 m/s². What is the horizontal component of the velocity?

10.336 m/s explanation The horizontal component of the velocity is vₓ= v₀cosθ

A block of mass 2.10317 kg lies on a friction-less table, pulled by another mass 4.12417 kg under the influence of Earth's gravity. The acceleration of gravity is 9.8 m/s². LOOK AT FIGURE 12 IN GOOGLE DOC What is the magnitude of the tension T of the rope between the two masses?

13.6501 N explanation Analyzing the horizontal forces on block m₁, we have ∑Fₓ : T= m₁a = (2.10317 kg) (6.49023 m/s²) = 13.6501 N

A place kicker must kick a football from a point 32 m (about 35 yd) from the goal, and the ball must clear the crossbar, which is 3.05 m high. When kicked, the ball leaves the ground with a speed of 24 m/s at an angle of 55° to the horizontal. By how much does the ball clear or fall short of clearing the crossbar? The acceleration of gravity is 9.8 m/s².

16.1724 m explanation Let: ∆x= 32 m v₀= 24 m/s θ= 55° hbar= 3.05 m and g= 9.8 m/s² From the horizontal motion ∆x= v₀cosθ*∆t since aₓ= 0 m/s², the time required for the ball to reach the position of the crossbar is ∆t = (∆x)/(v₀cosθ) At this time, the height of the football above the ground is ∆y= v₀sinθ*∆t-½g(∆t)² and the height above the bar is ∆h= ∆y- hbar = v₀sinθ * ((∆x)/(v₀cosθ))- ½g ((∆x)/(v₀cosθ))²- hbar = ∆x*tanθ- hbar - ((g(∆x)²/(2v₀²cos²θ)) = 32 m* tan55° - 3.05m - ((9.8 m/s²)(32 m)²)/((2(24 m/s)²*cos²55°)) =16.1724 m

LOOK AT FIGURE 10 IN GOOGLE DOC Given v₀=18.8 m/s and α= 30°, what is the speed of the projectile when it reaches its maximum height y=ymax; (i.e., at point A in the figure)?

16.2813 m/s explanation At its maximum height vᵧ= 0, thus the only component of velocity is in the x-direction. The speed is thus v=vₓ= v₀cosα = (18.8 m/s) cos30° = 16.2813 m/s

The elevators in the John Hancock building in Chicago move at 862 ft in 16 s. What is their speed?

16.4211 m/s explanation: v= 862 ft/16 s * (12 in/1 ft) * (2.54 cm/1 in) * (1 m/ 100 cm) = 16.4211 m/s

A book weighing 17 N is placed on a table. How much support force does a table exert on the book?

17 N explanation Newton's Third Law of Motion states that for every action, there is an equal but opposite reaction. The table exerts the same force up on the book as the book exerts downward on the table.

LOOK AT FIGURE 10 IN GOOGLE DOC Find the speed of the projectile, on its way down, at the height y=(ymax)/2 (i.e., at point B in the figure)

17.5858 m/s explanation At B, v₈= √(v²ₓ₈ + v²ᵧ₈) The horizontal component of velocity is constant, so vₓ₈= v₀ₓ= v₀cosα The trick is finding vᵧ₈. Applying v²=v₀²+2as for the y-component of motion from A to B gives v²ᵧ₈= 0 + 2g (h-(h/2)) = gh We need to determine h. Applying v²=v₀²+2as for the y-component of motion from O to A gives 0=v₀²ᵧ -2gh, or h= (v₀²ᵧ)/2g Substituting Eq. 3 into Eq. 2 v²ᵧ₈= g (v₀²ᵧ/2g) = v₀²ᵧ/2 Combining Eqs. 1 and 4 gives v₈= √(v₀ₓ²+v²ᵧ₈) = √((v₀cosα)² + ½(9.4 m/s)²) = 17.5858 m/s

A Knight of the Round Table fires off a vat of burning pitch from his catapult at 13.3 m/s, at 39⁰ above the horizontal. The acceleration of gravity is 9.8 m/s². How far from the catapult does it land?

17.6556 m explanation The horizontal velocity vₓ remains constant for the entire flight, so x= vₓt

You have two forces, 98.3 N and 71.8 N. What is their resultant (magnitude) if they both act downward?

170.1 N explanation Add forces acting in the same direction 98.3 N + 71.8 N = 170.1 N

A bee flies to a flower 586 m due south of its hive. The bee's speed in still air is 0.71 m/s, and there is a wind blowing toward the south at 0.19 m/s. How long will it take the bee to travel to the flower and back to the hive?

1778.03 s explanation The wind helps the bee fly faster on the downtrip than on the return trip. The time to go to the flower is given by twith = Swith/vwith = s/ (vbee + vwind) The wind slows the bee on the return trip, so the time for the return trip is tagainst = Sagainst/vagainst= s/ (vbee-vwind) The total time is t= twith + tagainst

A car is traveling at 53.4 mi/h on a horizontal highway. The acceleration of gravity is 9.8 m/s². If the coefficient of friction between road and tires on a rainy day is 0.15, what is the minimum distance in which the car will stop? (1 mi = 1.609 km)

193.751 m explanation Using that the frictional force acts in the opposite direction of the velocity and is given by, fₖ= µₖN and the normal force N = mg as the car is moving on a horizontal road, we apply Newton's second law in the direction of motion which gives -fₖ= -µₖ*mg= ma Solving for a a=-µₖg The acceleration a may be found from the following kinematics equation with v(final)= 0: v(final)²= v₀²+2ax a= (-v₀²)/(2x) Combining equations (1) and (2), we obtain (v₀²)/(2x)= µₖg or solving for x, x= (v₀²)/(2µₖg) Plugging in the appropriate values, x(wet)= (23.8669 m/s)²/((2(0.15)(9.8 m/s²))= 193.751 m

Consider the following graph of motion. LOOK AT FIGURE 1 ON GOOGLE DOC How far did the object travel between 2 s and 4 s? 1) 50 m 2) 20 m 3) 30 m 4) 40 m 5) 10 m

2) 20 m explanation: The particle moved from 40 m to 20 m, so Δd= 40 m - 20 m= 20 m

A 500-kilogram sports car accelerates uniformly from rest, reaching a speed of 30 meters per second in 6 seconds. What distance has the car traveled during the 6 seconds? 1) 30 m 2) 90 m 3) 15 m 4) 180 m 5) 60 m

2) 90 m explanation: Let: vf= 30 m/s and t=6 s Since v₀= 0 and s₀= 0, vf= v₀ + at= at a = (vf/t) s=s₀t + v₀t + 1/2at²= 1/2at²= 1/2 (vf/t)*t²= 1/2*vft = 1/2*(30 m/s)(6 s) = 90 m

Identify the action-reaction pairs when a base-ball is being hit 1) Action: bat hits ball Reaction: ball hits the player 2) Action: bat hits ball Reaction: ball hits bat 3) Action: ball hits bat Reaction: bat hits the player 4) None of these 5) Action: ball hits bat Reaction: bat pushes the arm of the player

2) Action: bat hits ball Reaction: ball hits bat explanation As the bat hits the ball, the ball also hits the bat; this forms an action-reaction pair

A child throws a steel ball straight up. Consider the motion of the ball after it has left the child's hand but before it touches the ground, and assume that forces exerted by the air are negligible. For these conditions, identify the force(s) acting on the ball. 1) An almost constant downward force of gravity along with an upward force that steadily decreases until the ball reaches its highest point; on the way down only an almost constant force of gravity 2) An almost constant downward force of gravity only 3) None of these; the ball falls back to the ground because of its natural tendency to rest on the surface of the earth 4) A downward force of gravity along with a steadily decreasing upward force 5) A steadily decreasing upward force from the moment it leaves the child's hand until it reaches its highest point; on the way down, a steadily increasing downward force of gravity as the object gets closer to earth

2) An almost constant downward force of gravity only explanation: Once the child lets go of the ball, no contact forces act on the ball. The only force acting on the ball is the force of gravity, directed downward, and essentially constant because the ball travels a small distance compared to the diameter of the earth

A toy car is given a quick push so that is moves up an inclined ramp. After it is released, it moves up, reaches its highest point and moves back down again. Friction is so small it can be ignored. LOOK AT FIGURE 11 ON GOOGLE DOC What net force acts on the car? 1) Net increasing force up the ramp 2) Net constant force down the ramp 3) Net increasing force down the ramp 4) Net decreasing force up the ramp 5) Net force of zero 6) Net constant force up the ramp 7) Net decreasing force down the ramp

2) Net constant force down the ramp explanation The net force is the component of gravity along the incline; it is constant and acts down the ramp

Consider a book that remains at rest on an incline. LOOK AT FIGURE 18 ON GOOGLE DOCS Compare the normal force exerted on the book by the inclined plane and the weight force exerted on the book by the earth. Are they equal in magnitude? 1) Yes 2) No 3) Their magnitudes cannot be determined since the forces are not in the same direction

2) No explanation They are not equal in magnitude. The normal force given by N=mgcosθ, with θ the angle the incline makes with the ground. Since |cosθ| is less than 1 as long as the incline is not horizontal, the magnitude N of the normal force, will be less than the magnitude mg of the weight

A space probe is carried by a rocket into outer space where it continues to move on its own in a straight like. What keeps the probe moving? 1) A propeller 2) Nothing specific; in the absence of forces it would continue moving in a straight line 3) Nothing; the probe will eventually stop 4) The gravitation forces from different stars and planets 5) None of these

2) Nothing specific; in the absence of forces it would continue moving in a straight line explanation According to the law of inertia, every object at rest remains at rest, or remains in motion with constant velocity in a straight line unless external forces act on it; so inertia keeps the probe moving in outer space

If you drop a rock from a height of 17 m, it accelerates at g and strikes the ground 1.86263 s later. If you drop the same rock from half the height, what will its acceleration be? The acceleration of gravity is 9.8 m/s². 1) 0 2) The same 3) About half 4) Unable to determine 5) More

2) The same explanation: The acceleration of an object due to gravity is virtually constant at the earth's surface and is not affected by height

Find the dimensions for the quantity of c in the expression: s=ct² 1) [c]=[L] 2) [c]=[L]/[T²] 3) [c]=1/[T] 4) [c]=[T]/[L] 5) [c]=[L][T] 6) [c]=[L][T²] 7) [c]=[L²] 8) [c]=1/[L] 9) [c]=[L]/[T] 10) [c]=[T]

2) [c]=[L]/[T²] explanation: c=s/t² [c]= [s]/[t²]= [L]/[T²]

You have two forces, 98.3 N and 71.8 N. What is the direction of this resultant? 1) upward 2) downward

2) downward explanation The direction of the resultant is that of the original forces

Consider the expression 2π√(l/g), where l is length and g is gravitational acceleration in units of length divided by time squared. Evaluate its units. 1) s² 2) s 3) m 4) (s/m)² 5) m² 6) s/m 7) m/s 8) (m/s)²

2) s explanation: [l]=m and [g]= m/s² 2 and π are constants, so √[l]/[g] = √m/ (m/s²) = √m* (s²/m) = √s² = s

Length is to meter as 1) weight is to mass 2) temperature is to celsius 3) density is to volume 4) liter is to distance

2) temperature is to celsius - the standard metric unit for length is the meter, and the standard metric unit for temperature is the celsius

A block of mass m is accelerated across a rough surface by a force of magnitude F that is exerted at an angle θ with the horizontal, as shown above. The frictional force on the block exerted by the surface has magnitude f. LOOK AT FIGURE 20 ON GOOGLE DOC What is the magnitude of the acceleration (a) of the block? 1) |a|= (Fsinθ -mg)/m 2) |a|= (Fcosθ - f)/m 3) |a|= (Fcosθ)/m 4) |a|= (F-f)/m 5) |a|= F/m

2) |a|= (Fcosθ - f)/m explanation From Newton's second law of motion, the acceleration is the total force in the horizontal direction divided by the mass. There are two forces in the horizontal direction: one is the friction force; the other is the horizontal component of the dragging force F, but they are in the opposite directions, so the acceleration of the block is |a|= (Fcosθ - f)/m

The barrel of a rifle has a length of 0.908 m. A bullet leaves the muzzle of a riffle barrel with a speed of 619 m/s. What is the acceleration of the bullet while in the barrel? A bullet in a riffle barrel does not have a constant acceleration, but constant acceleration is to be assumed for this problem.

2.10992 x 10⁵ m/s² explanation: Let: l = 0.908 m v₀ = 0 m/s vf= 619 m/s Under constant acceleration, vf² = v₀² +2al = 2al a= vf²/2l = (619 m/s)²/(2*0.908 m)= 2.10992 x 10⁵ m/s²

Both car A and car B leave school at the same time, traveling in the same direction. Car A travels at a constant speed of 79 km/h, while car B travels at a constant speed of 88 km/h. How long does it take car B to reach the gas station?

2.42045 h explanation: t= d/v₂= (213 km/88km/h) = 2.42045 h

A cat chases a mouse across a 1.4 m high table. The mouse steps out of the way, and the cat slides off the table and strikes the floor 1.4 m from the edge of the table. What was the cat's speed when it slid off the table? The acceleration of gravity is 9.81 m/s²

2.6205 m/s explanation ∆x= 1.4 m, ∆y= 1.4 m, and g= 9.81 m/s² Horizontally, aₓ= 0 m/s² ∆x=v₀ₓ∆t-½g(∆t)²= -½g(∆t)² ∆t= √((2∆y)/(-g))= (∆x)/(vₓ) vₓ= √((-g)/(2∆y)) *∆x = √((-9.81 m/s²)/2(-1.4 m) * (1.4 m) = 2.6205 m/s

2.6962 h explanation: t= d/v₁= (213 km/79km/h) = 2.6962 h

A(n) 85 kg block is pushed along the ceiling with a constant applied force of 1100 N that acts at an angle of 70° with the horizontal, as in the figure. The block accelerates to the right at 3.8 m/s² LOOK AT FIGURE 22 ON GOOGLE DOC The acceleration of gravity is 9.8 m/s². What is the magnitude of the normal force the ceiling exerts on the block?

200.662 N explanation Given: θ= 70° a= 3.8 m/s² and m=85 kg and F= 1100 N Take up to be the positive y direction and to-the-right to be the positive x direction. Applying Newton's second law in the x and y directions, we have y: Fsinθ-mg-N= 0 x: Fcosθ-µN= ma Solving equation 1 for N, we have N= Fsinθ-mg = (1100 N) sin70°- (85 kg)*9.8 m/s² =200.662 N which is positive since the normal force in Eq. 1 has a minus sign in front of it because it is pointing downward

Convert 47.5 mi/h to m/s. 1 mi=1609 m

21.2299 m/s explanation: 1 mi = 1609 n 1 h= 3600 s (47.5 mi/h) * (1609m/1 mi) *(1 h/3600 s) = 21.2200 m/s

Both car A and car B leave school at the same time, traveling in the same direction. Car A travels at a constant speed of 79 km/h, while car B travels at a constant speed of 88 km/h. How far is car B from school 2.4 h later?

211.2 km explanation: d=v₂t= (88 km/h) (2.4 h)= 211.2 km

Express the length of a 240 yd dash in meters.

219.456 m explanation: Since 1 yd= 0.9144 m, l= (240 yd) *(0.9144 m/1 yd) = 219.456 m

What is the stopping distance when the surface is dry and the coefficient of friction is 0.536?

22.2986 m explanation Given: µₖ= 0.536 For a dry surface, ∆x= (v²₀)/(2µₖg) = ((55.1 km/h)² * (1000 m/1 km)² * (1 h/3600 s)²)/(2*(0.536)*(9.8 m/s²)) = 22.2986 m

You have two forces, 98.3 N and 71.8 N. What is their resultant (magnitude) if the first acts upward and the second acts downward?

26.5 N explanation Subtract forces acting in opposite directions: 98.3N -71.8 N= 26.5 N

When you jump vertically off the ground, what is your acceleration when you reach your highest point? Up is positive. 1) g 2) 0 m/s² 3) -g 4) g/2 5) g/3 6) -g/3 7) -g/2 8) All are wrong

3) -g explanation At the top of your jump your acceleration is still -g. Let the equation for acceleration via Newton's second law guide your thinking. a= (F/m) = -mg/m= -g Gravity does not cease to act at any point of your jump. The acceleration of gravity is directed towards the center of the Earth.

What is the speed from 2 s to 4 s? LOOK AT FIGURE 1 ON GOOGLE DOC 1) 0 m/s 2) 20 m/s 3) 10 m/s 4) 15 m/s 5) 5 m/s

3) 10 m/s explanation: v= Δd/Δt = (40 m - 20 m)/2 s= 10 m/s

Which conversion factor would you use to change 18 kilometers to meters? 1) 1 km/ 100 m 2) 1 km/ 1000 m 3) 1000 m/ 1 km 4) 100 m/ 1 km

3) 1000 m/ 1 km explanation: There are 1000 meters in every kilometer. Since you are trying to get rid of km, the conversion is (18 km) *1000 m/ 1 km

Consider a book that remains at rest on an incline. LOOK AT FIGURE 18 ON GOOGLE DOCS Are they opposite in direction? 1) No; the normal force acts up the incline to keep the book from sliding down. 2) Yes; the normal force always acts opposite the weight force. Otherwise, the book would fall through the inclined plane. 3) No; the normal force acts perpendicular to the surface of the inclined plane. 4) Yes; the normal force acts opposite to the weight force because the book is stationary.

3) No; the normal force acts perpendicular to the surface of the inclined plane. explanation The normal force acts perpendicular to the surface of the inclined plane. Because the plane is inclined, this is not opposite in direction to the weight force.

Find the dimensions for the quantity c in the expression: s=r*cos(ct) 1) [c]=1/[L] 2) [c]=[T] 3) [c]=1/[T] 4) [c]=[L²] 5) [c]=[L]/[T] 6) [c]=[L][T²] 7) [c]=[L]/[T²] 8) [c]=[L][T] 9) [c]=[L] 10) [c]=[T]/[L]

3) [c]=1/[T] explanation: the argument of a trig function is dimensionless (radians), so ct=radians [c]= rad/[t]= 1/[T]

In the absence of a gravitational field, you could determine the mass of an object (of unknown composition) by 1) measuring its volume 2) weighing it 3) applying a known force and measuring its acceleration

3) applying a known force and measuring its acceleration explanation In Newton's second law (F=ma), mass is the quantity that relates force and acceleration. If both force and acceleration are known, the mass can be calculated.

SEE FIGURE 6 ON GOOGLE DOC A boy throws a ball upward. Compare the magnitudes of the gravitational accelerations at three points along the path of the ball. Point A before the ball reaches the top. a) Point A on the way up b) Point B is at the top c) Point C is after it has passed the top and on the way down The magnitudes of acceleration are related as 1) a₈ < g (but a₈≠0) 2) a₈ > g 3) a₈ = g 4) a₈ = 0

3) a₈ = g explanation The gravitational acceleration near the surface of the earth is a constant, so the correct choice is a₈ = g

SEE FIGURE 6 ON GOOGLE DOC A boy throws a ball upward. Compare the magnitudes of the gravitational accelerations at three points along the path of the ball. Point A before the ball reaches the top. a) Point A on the way up b) Point B is at the top c) Point C is after it has passed the top and on the way down The magnitudes of the acceleration are related as 1) aₐ > a₈ 2) aₐ < a₈ 3) aₐ = a₈

3) aₐ = a₈ explanation The gravitational acceleration near the surface of the earth is a constant, so the correct choice is aₐ = a₈

The graph indicates LOOK AT FIGURE 1 ON GOOGLE DOC 1) no motion 2) decreasing velocity 3) constant velocity 4) increasing velocity 5) constant position

3) constant velocity explanation: The slope of the graph is the same everywhere, so the graph indicates constant positive velocity

The force that opposes motion is 1) falling 2) inertia 3) friction 4) unbalanced

3) friction

A ball is thrown upward with an initial vertical speed of v₀ to a maximum height of hmax. LOOK AT FIGURE 4 ON GOOGLE DOC What is its maximum height? The acceleration of gravity is g. Neglect air resistance. 1) hmax = (v₀²)/(√2*g) 2) hmax = (√5 *v₀²)/(2√2 * g) 3) hmax = (v₀²)/2g 4) hmax = (√3 * v₀²)/(2√2 *g) 5) hmax = v₀²/g 6) hmax = (5*v₀²)/(8g) 7) hmax = v₀²/4g 8) hmax = (√3 *v₀²)/2g 9) hmax = (3*v₀²)/4g

3) hmax = (v₀²)/2g explanation: With up positive, a= -g and for the upward motion, tf = 0, so for constant acceleration, vf² = v₀² + 2a (x-x₀) 0= v₀² -2g (hmax-0) hmax= v₀²/2g

Which of the following expressions for the coefficient of friction is correct? 1) µ = (mg)/f 2) µ= f/ (mg-Fcosθ) 3) µ= f/ (mg-Fsinθ) 4) µ= (mg-Fcosθ)/f 5) µ= f/(mg)

3) µ= f/ (mg-Fsinθ) explanation By definition, the coefficient of kinetic friction is the ratio of the friction force and the normal force in the vertical direction. And it is easy to see that the normal force in the vertical direction is just N=mg- Fsinθ so the coefficient of friction is µ= f/ (mg-Fsinθ)

The position of a softball tossed vertically upward is described by the equation y=c₁t-c₂t² where y is in meters, t in seconds, c₁= 3.39 m/s, and c₂= 3.04 m/s². Find the ball's initial speed v₀ at t₀= 0s

3.39 m/s explanation: Let: t=0 The velocity is simply the derivative of y with respect to t: v= (dy/dt) = 3.39 m/s - 2(3.04 m/s²)t, so v₀= 3.39 m/s

What is 16.4211 m/s in miles per hour?

36.7341 mi/h explanation: Since 1 mi= 5280 ft and 1 h= 3600 s v= (862 ft/16 s) * (3600 s/1 h) * (1 mi/ 5280 ft) = 36.7341 mi/h

How much force must be applied during liftoff to accelerate a 37 kg satellite just enough to counter the Earth's gravitational acceleration of 9.8 m/s²?

362.6 N explanation Let: m=37 kg and g=9.8 m/s² In order to lift the satellite, a force equal to the weight of the satellite must be supplied: F=mg = (37 kg) (9.8 m/s²) = 362.6 N

A passenger at the rear of a train traveling at 17 m/s relative to Earth throws a baseball with a speed of 13 m/s in the direction opposite of motion of the train. What is the velocity of the baseball relative to Earth as it leaves the thrower's hand?

4 m/s explanation v₈ₑ= v₈t + vₑt Given: vtₑ= 17 m/s v₈t= -13 m/s v₈ₑ= -13 m/s + 17 m/s = 4 m/s

What is the standard form for 2 x 10^-3? 1) 0.0002 2) 0.2 3) 2000 4) 0.002 5) 20 6) 200 7) 2 8) 0.02

4) 0.002 explanation: The exponent is negative, so the number is less than 1. The decimal is to be moved 3 places, so the number is 0.002.

Despite a very strong wind, a tennis player manages to hit a tennis ball with her racquet so that the ball passes over the net and lands in her opponent's court. Consider the following forces: 1. A downward force of gravity, 2. A force by the hit, and 3. A force exerted by the air. Which of the above forces is (are) acting on the tennis ball after it has left contact with the racquet and before it touches the ground? 1) 1 and 2 2) 1 only 3) 2 and 3 4) 1 and 3 5) 1, 2, and 3

4) 1 and 3 explanation The forces acting on the tennis ball after it has left contact with the racquet are gravity and air resistance. A force by the hit is no longer acting on the ball.

What is the speed of Swimmer 2? LOOK AT FIGURE 3 ON GOOGLE DOC 1) 8 m/s 2) 6 m/s 3) 3 m/s 4) 1 m/s 5) 5 m/s 6) 7 m/s 7) 2 m/s 8) 4 m/s 9) None of these

4) 1 m/s explanation: speed = distance/time = 20 m/20 s = 1 m/s

A heavily loaded freight train moves with constant velocity. What is the relationship between the net force on the first car (F₁) and the net force on the last car (F₂)? 1) F₁ < F₂ 2) F₁ > F₂ 3) Unable to determine 4) F₁ = F₂

4) F₁ = F₂ explanation The net force on each car is zero because the train moves with constant velocity (no acceleration)

The volume of an object is given as a function of time by V=A + B/t +Ct⁴ Find the dimension of the constant C. 1) L⁴/T³ 2) L/T 3) L²/T⁴ 4) L³/T⁴ 5) L/T⁴

4) L³/T⁴ explanation: [Ct⁴] = [V] [C] T⁴ = L³ C= L³/T⁴

Which swimmer has the greatest speed? LOOK AT FIGURE 3 ON GOOGLE DOC 1) Unable to determine 2) Swimmer 2 3) The speeds are the same 4) Swimmer 1

4) Swimmer 1 explanation: Swimmer 1 has the steeper graph, which means the greater speed

A certain force with units of measure (ML)/T² is given by the equation F= (KML²)/T⁴ where M is a mass, L is a length, T is a time, and K is a constant. What are the units of the constant K? 1) [K]= (M²L²)/T⁶ 2) [K]= (ML)/T² 3) [K]= L/T² 4) [K]= T²/L 5) [K]= T⁶/(M²L²) 6) [K]= T²/(ML)

4) [K]= T²/L explanation Solving for K, K= (FT⁴)/(ML²) Since the dimensions of force are [F]= (ML)/T², we get [K]= (((ML)/T²)*T⁴)/(ML²)= T²/L

No force is necessary to 1) stop an object from moving 2) cause a change in the motion of an object 3) start an object moving 4) keep an object moving the way it is already moving

4) keep an object moving the way it is already moving explanation bc

One Newton is the force 1) that gives 1 kg body an acceleration of 9.8 m/s² 2) of gravity on a 1 kg body 3) of gravity on a 1 g body 4) that gives a 1 kg body an acceleration of 1 m/s² 5) that gives a 1 g body an acceleration of 1 cm/s²

4) that gives a 1 kg body an acceleration of 1 m/s² explanation One Newton is a kg*m/sec², so such a force acting on a 1 kg body gives a 1 m/sec² acceleration

The "reaction" force does not cancel the "acceleration" force because: 1) the action force is greater than the reaction force 2) they are in the same direction 3) the reaction is greater than the acceleration force 4) they act on different bodies 5) the reaction force exists only after the action force is removed

4) they act on different bodies explanation The statement of Newton's Third Law is that the force exerted by one body on another will inevitably cause that body to exert an equal and opposite force on the first object. Therefore, the two forces act on two different objects, and cannot cancel.

A 28 kg block slides down a frictionless slope which is at angle θ=31°. Starting from rest, the time to slide down is t=1.92 s The acceleration of gravity is 9.8 m/s². LOOK AT FIGURE 13 IN GOOGLE DOC What is the total vertical height through which the block descended?

4.79157 m explanation The vertical height is found from trigonometry h=s * sinθ = (9.30332 m) x sin31° = 4.79157 m

A rocket sled for testing equipment under large accelerations starts at rest and accelerates according to the expression a= (2.7 m/s³) t + (7 m/s²) How far does the sled move in the time interval t=0 to t=2.9 s?

40.41 m explanation: Let: α= 2.7 m/s³ β= 7 m/s² and t= 2.9 s a= (dv/dt)= αt+ β, so v= ∫(αt' +β) dt'= (α/2) * t² + βt= (dx/dt) x= ∫((α/2)* t'² +βt') dt' = (α/6)t³ +(β/2)t² = ( 2.7 m/s³ /6)*(2.9 s)³ + (7 m/s² /2) *(2.9 s)² = 40.41 m

40.4169 N explanation m₁= 2.10317 kg and m₂= 4.12417 kg The net force on the system is simply the weight of m₂ Fnet= m₂g = (4.12417 kg) (9.8 m/s²) = 40.4169 N

Express the length of a marathon run of 26 mi 385 yd in meters.

42186 m explanation: Since 1 mi= 1609 m and 1 yd= 0.9144 m l = (26 mi) (1609 m/1 mi) + (385 yd) (0.9144 m/1 yd) = 42186 m

A car is traveling at 53.4 mi/h on a horizontal highway. The acceleration of gravity is 9.8 m/s². What is the stopping distance when the surrounding surface is dry and µ(dry)=0.647?

44.919 m explanation x(dry)= (23.8669 m/s)²/((2*(0.647)(9.8 m/s²)) = 44.919 m (a considerable difference)

Consider two vectors F₁= 77 N, F₂= 83 N, θ₁=226⁰, and θ₂= 280⁰, measured from the positive x axis with the counter-clockwise angular direction as positive. What is the magnitude of the resultant vector? F=F₁+F₂ 1) 178.841 N 2) 173.42 N 3) 166.066 N 4) 169.093 N 5) 142.587 N 6) 127.734 N 7) 93.8832 N 8) 128.304 N

5) 142.587 N explanation: *Draw it to scale The components of the forces are Fₓ= Fcosθ and Fᵧ= Fsinθ, so F₁ₓ= (77 N) cos 226⁰= -55.4887 N F₂ₓ= (83 N) cos 280⁰= 14.4128 N F₁ᵧ= (77 N) sin 226⁰ = -55.3892 N F₂ᵧ= (83 N) sin 280⁰ = -81.739 N The components of the resultant vectors are Fₓ= F₁ₓ+ F₂ₓ = -55.4887 N + 14.4128 N = -39.0759 N and Fᵧ= F₁ᵧ + F₂ᵧ = -55.3892 N + -81.739 N = -137.128 N, so the magnitude of the resultant vector is ‖F‖= √(Fₓ²+Fᵧ²) = √((-39.0759 N)² + (-137.128 N)²) = 142.587 N

The tension of the strings are T₁ and T₂ (see sketch). The equation of motion of m₂ is given by 1) T₁= m₂a 2) T₁ + T₂ = (m₁ + m₃) a 3) T₁ + T₂ = m₁a 4) T₁ + T₂ = m₂a 5) T₁ - T₂ = m₂a 6) T₁ - T₂ = (m₁ + m₃) a 7) T₁ - T₂ = m₁a 8) T₁ = m₁a 9) T₁ = (m₁ + m₃) a

5) T₁ - T₂ = m₂a explanation By inspection, the net force acting on m₂ is T₁ - T₂, so the equation of motion is given by T₁ - T₂= m₂a

SEE FIGURE 6 ON GOOGLE DOC A boy throws a ball upward. Compare the magnitudes of the gravitational accelerations at three points along the path of the ball. Point A before the ball reaches the top. a) Point A on the way up b) Point B is at the top c) Point C is after it has passed the top and on the way down The magnitudes of the acceleration are related as 1) a꜀ < aₐ and a꜀ < a₈ 2) a꜀ > aₐ and a꜀ > a₈ 3) a꜀ < aₐ and a꜀ > a₈ 4) a꜀ > aₐ and a꜀ < a₈ 5) a꜀ = aₐ = a₈

5) a꜀ = aₐ = a₈ explanation The gravitational acceleration near the surface of the earth is constant, so the correct choice is a꜀ = aₐ = a₈

The graph indicates LOOK AT FIGURE 2 ON GOOGLE DOC 1) no motion 2) constant position 3) constant velocity 4) decreasing velocity 5) increasing velocity

5) increasing velocity explanation: The slopes are steeper as time goes on, so the velocities are increasing

The position x has the dimensions of [L] (i.e., length). The velocity v has the dimensions of [L]/[T] (i.e., length/time). The acceleration a has the dimensions of [L]/[T]². Which of the following quantities has the same dimensions as a? 1) xv/a 2) x²/v 3) v² 4) a²/v 5) v²/x

5) v²/x explanation: The answer is v²/x, which has dimensions [v²/x] = (L/T)² (1/L) = L/T² = [a] None of the other choices have dimensions that work out.

Express the distance 3273 mi from Seattle, Washington, to Miami, Florida, in meters.

5.26626 * 10⁶ m explanation: Since 1 mi= 1609 m, l= (3273 mi) * (1609 m/1 mi) = 5.26626 * 10⁶ m

A 4 kg block rests on a horizontal table, attached to a 2 kg block by a light string as shown in the figure. The acceleration of gravity is 9.81 m/s². LOOK AT FIGURE 21 ON GOOGLE DOC If the coefficient of static friction is less than that found above, and if the coefficient of kinetic friction between the block and the table is 0.4, find the time it takes for the 2 kg mass to fall 10 m to the floor if the system starts from rest.

5.53001 s explanation Let: µₖ = 0.4 and ∆x= 10 m Applying ∑Fₓ= m₁aₓ to box 1 and ∑Fᵧ= m₂aᵧ to box 2, T-µₖ *m₁g= m₁a m₂g-T=m₂a Adding the equations, (m₂-µₖ *m₁)g/m₁+m₂ = [2 kg-0.4(4 kg)] (9.81 m/s²)/(4 kg+ 2 kg) =0.654 m/s² Applying kinematics to the system starting from rest, ∆x=v₀∆t +½a (∆t)² = ½a(∆t)² ∆t = √((2∆x)/a)) = √((2 (10m)/(0.654 m/s²)) =5.53001 s

A physics book is moved once around the perimeter of a table of dimensions 1 m by 3 m. If the book ends up at its initial position, what is the magnitude of its displacement? 1) 1 m 2) 7 m 3) 2 m 4) 3 m 5) 6 m 6) 0 m 7) 8 m 8) 4 m 9) 5 m 10) None of these

6) 0 m explanation: Since its final position is the same as its initial position, the displacement is 0

A building superintendent twirls a set of keys in a circle at the end of a cord. If the keys have a centripetal acceleration of 132 m/s² and the cord has a length of 0.31 m, what is the tangential speed of the keys?

6.39687 m/s explanation a꜀= (vt²)/r a꜀= 132 m/s² r= 0.31 m vt=√(ra꜀) =√((0.31 m)(132 m/s²)) = 6.39687 m/s

6.49023 m/s² explanation From Netwon's Second Law, Fnet= m₂g= (m₁ + m₂) a Solving for a, a= ((m₂)/(m₁ + m₂)) *g = ((4.12417 kg)/(2.10317 kg + 4.12417 kg)) * (9.8 m/s²) = 6.49023 m/s²

A dog pulls on a pillow with a force of 7.4 N at an angle of 20.8° above the horizontal a) What is the x component of this force?

6.91771 N explanation Fₓ= Fcosθ F=7.4 N θ= 20.8° Fₓ= (7.4 N) cos20.8° = 6.91771 N

What is the direction of the resultant vector F? Use counterclockwise as the positive angular direction, between the limits of -180⁰ and +180⁰ from the positive x axis. 1) 50.00369⁰ 2) 133.766⁰ 3) 158.881⁰ 4) -23.9575⁰ 5) 31.462⁰ 6) -97.9837⁰ 7) -105.905⁰ 8) -120.208⁰

7) -105.905⁰ explanation: The angle is θ= arctan (Fᵧ/Fₓ) =tan^-¹((-137.128 N)/(-39.0759 N)) = -105.905⁰ The arctan function is defined between the limits of -90⁰ and +90⁰, so you must check for the quadrant in which your resultant vector lies

Predict the distance Swimmer 1 can go in 60 seconds. LOOK AT FIGURE 3 ON GOOGLE DOC 1) 50 m 2) 40 m 3) 60 m 4) 130 m 5) 100 m 6) 110 m 7) 120 m 8) 70 m 9) 80 m 10) 90 m

7) 120 m explanation: There is an increase of 20 m every 10 s, so the distance would be 120 m

What is the speed of Swimmer 1? LOOK AT FIGURE 3 ON GOOGLE DOC 1) 4 m/s 2) 8 m/s 3) 1 m/s 4) 7 m/s 5) 3 m/s 6) 5 m/s 7) 2 m/s 8) 6 m/s 9) None of these

7) 2 m/s explanation: speed = distance/time = 40 m/20 s = 2 m/s

Consider the following graph of motion LOOK AT FIGURE 2 ON GOOGLE DOC How far did the object travel between 3 s and 9 s? 1) 200 m 2) 500 m 3) 50 m 4) 450 m 5) 100 m 6) 250 m 7) 350 m 8) 150 m 9) 400 m 10) 300 m

7) 350 m explanation: The particle moved from 50 m to 400 m, so the distance Δd = 400 m - 50 m= 350 m

A projectile is fired with an initial speed v₀ at t=0. The angle between the initial velocity v₀ and the horizontal plane is α. LOOK AT FIGURE 10 IN GOOGLE DOC The time tmax it takes for the projectile to reach its maximum height is 1) tmax= (v₀sinα)/2g 2) tmax= (v₀cosα)/g 3) tmax= (v₀)/g 4) tmax= g/(v₀) 5) tmax= v₀ 6) tmax= v₀g 7) tmax= (v₀sinα)/g 8) tmax= (v₀cosα)/2g

7) tmax= (v₀sinα)/g explanation For two dimensional projectile motion in a gravitational field the acceleration is due to gravity only and acts exclusively on the y component of velocity. Kinematic equations for constant acceleration are applicable. Velocity in the x-direction is a constant. The initial y-velocity is vᵧ₀= v₀sinα. The projectile reaches its maximum height when vᵧ=0 Under this condition the basic equation v=v₀ +at, reduces at vᵧ=0=vᵧ₀-gt, so tmax= vᵧ₀/g= (v₀sinα)/g

An object is in free fall. At one instant, it is traveling at 57 m/s. Exactly 1.9 s later, what is its speed? The acceleration of gravity is 9.8 m/s².

75.62 m/s explanation: Let: g=9.8 m/s² v₀= 57 m/s and t= 1.9 s Since all the motion is down, we can consider down to be positive. The speed of an object in free-fall is given by vf= v₀ + gt= 57 m/s + (9.8 m/s²) (1.9 s) = 75.62 m/s

An object has an acceleration given by aₓ(t)= Dt². Its initial position is x(0)= A and its initial velocity vₓ(0)= B where A, B, and D are constants. For this object, what is x(t)? 1) A + Bt + 1/2* Dt² 2) (D+B)*(t/2)² 3) A + Bt + 1/3*Dt⁴ 4) At+ Bt+ 1/6*Dt⁴ 5) A + Bt + 1/4*Dt⁴ 6) At + Bt² + 1/4*Dt⁴ 7) At + Bt² + 1/12 * Dt³ 8) A + Bt + 1/12* Dt⁴

8) A + Bt + 1/12* Dt⁴ explanation: Since aₓ = dvₓ/dt ∫dvₓ = ∫ aₓ(t) dt= (Dt³)/3 = vₓ(t)- vₓ(0) = vₓ(t)-B vₓ(t)= B +1/3*Dt³

Consider the three force vectors, F₁ with manitude 91 N and direction 120⁰, F₂ with magnitude 86 N and direction 201⁰, and F₃ with magnitude 58 N and direction 1⁰. All direction angles θ are measured from the positive x axis: counter-clockwise for θ>0 and clock-wise for θ<0. What is the magnitude of the resultant vector ‖F‖, where F= F₁ +F₂ + F₃?

83.651 N explanation: Let: F₁= 91 N and θ₁= 120⁰ F₂= 86 N and θ₂= 201⁰ F₃= 58 N and θ₃= 1⁰ Each force vector is defined by components Fₓ= F₁ₓ+ F₂ₓ + F₃ₓ = (91 N) cos120⁰ + (86 N) cos201⁰ + (58 N) cos1⁰ = -67.7968 N and Fᵧ= F₁ᵧ + F₂ᵧ + F₃ᵧ = (91 N) sin120⁰ + (86 N) sin201⁰ + (58 N) sin1⁰ = 49.0009 N with a magnitude of ‖F‖= √(Fₓ²+Fᵧ²) = √((-67.7968 N)² + (49.0009 N)²) = 83.651 N

A car is traveling at 55.1 km/h on a flat highway. The acceleration of gravity is 9.8 m/s². If the coefficient of friction between the road and the tires on a rainy day is 0.137, what is the minimum distance in which the car will stop?

87.2412 m explanation v₀= 55.1 km/h and µₖ= 0.137 Horizontally, Fₓnet= ma = -fₖ= -µₖN so that a= (-µₖN)/m= (-µₖm*g)/m= -µₖg v²(final)= v²₀ + 2a∆x=0 since v(final)= 0 m/s, so ∆x= (-v²₀)/2a = (-v²₀)/2*(-µₖg) = ((55.1 km/h)²*(1000 m/1 km)² *(1 h/3600 s)²)/(2*(0.137)*(9.8 m/s²)) = 87.2412 m

What is the average speed from 3 s to 9 s? LOOK AT FIGURE 2 ON GOOGLE DOC 1) 60 m/s 2) 30 m/s 3) 40 m/s 4) 47 m/s 5) 20 m/s 6) 36 m/s 7) 25 m/s 8) 50 m/s 9) 58 m/s

9) 58 m/s explanation: v= (Δd/Δt) = (400 m - 50 m)/6 s= 58 m/s

LOOK AT FIGURE 4 ON GOOGLE DOC Find the speed of the ball (in terms of the initial speed v₀) as the ball passes a point A (expressed as V₁ here), at one quarter of the maximum height. 1) V₁ = v₀/4 2) V₁ = v₀/2 3) V₁ = (√5* v₀) / (2√2) 4) V₁ = v₀/√2 5) V₁ = 3*v₀/4 6) V₁ = 5*v₀/8 7) V₁ = (√3*v₀)/(2√2) 8) V₁ = 3v₀/8 9) V₁ = (√3 *v₀) /2

9) V₁ = (√3 *v₀) /2 explanation: Δh = hmax/4= v₀²/8g, so vf²= v₀² + 2a (y-y₀) V₁²= v₀² -2g (v₀²/8g) = 3/4 v₀² V₁ = (√3/2) v₀

9.30332 m explanation Given m=28 kg and θ= 31° Consider the FBDs Fₓnet= Fcosθ- F₉‖= 0 F₉‖= mgsinθ= 0 Align the x axis along the incline. The weight component acting down the plane is mgsinθ, so Newton's Second Law gives mgsinθ= ma Solving for a, a= (9.8 m/s²) sin31° Using x-x₀= v₀t + ½ at² with x-x₀= s and v₀= 0, sp s=0.5at² = 0.5 * (5.04737 m/s²) (1.92 s)² = 9.30332 m

Block one is resting on top of block two, and the system of blocks is accelerating to the right due to a force applied to block two. LOOK AT FIGURE 17 ON GOOGLE DOCS In what direction is the frictional force being exerted on the upper block (block one), and is it kinetic friction or static friction?

Static friction to the right explanation The only force acting on the upper block in the horizontal direction is the force of static friction exerted by the lower block; that is the only force available to accelerate block one along with block two.

A mass slides with negligible friction on a circular track of 1 m radius oriented vertically. Its speed at the position shown in the figure is 3.13 m/s. LOOK AT FIGURE 8 ON GOOGLE DOC At the position shown in the figure, describe an arrow that would best represent the direction of the acceleration of the mass? The acceleration of gravity is 9.8 m/s² and no external forces act on the system.

The arrow would be pointing down and to the left 45° explanation The magnitude of the centipetal acceleration is aᵣ= (v²/r) = (3.13 m/s)²/1 m ≈ 9.8 m/s², acting inward with gravity acting down, thus the resultant vector being southwest

Consider a book that remains at rest on an incline. LOOK AT FIGURE 18 ON GOOGLE DOCS Draw a FBD for the book.

The frictional force keeps the book from sliding down. The normal force acts perpendicular to the inclined surface. The gravitational force acts down.

An automobile moves at a constant speed down one hill and up another along a smoothly curved surface as shown below. LOOK AT FIGURE 9 IN GOOGLE DOC Describe the velocity and acceleration arrows of the automobile at the instant that is the lower position as shown.

The velocity arrow would be pointing right, and the acceleration arrow would be pointing straight up explanation At the lowest point, the instantaneous velocity of the automobile is directed horizontally to the right. Since the speed of the automobile is unchanged, there is no tangential acceleration. Also since the path of the automobile is curved, there is centripetal acceleration, pointing upward.

The position of a softball tossed vertically upward is described by the equation y=c₁t-c₂t² where y is in meters, t in seconds, c₁= 3.39 m/s, and c₂= 3.04 m/s². Find its acceleration at t=1.03 s

-6.08 m/s² explanation: a=dv/dt= -2(3.04 m/s²) = -6.08 m/s²

A car accelerates from rest at 5.9 m/s². How much time does it need to attain a speed of 7 m/s?

1.18644 s explanation: Let: a= 5.9 m/s² and v= 7 m/s vf= v₀ + at = at t = vf/a = (7 m/s)/(5.9 m/s²) = 1.18644 s

A Knight of the Round Table fires off a vat of burning pitch from his catapult at 13.3 m/s, at 39⁰ above the horizontal. The acceleration of gravity is 9.8 m/s². How long is it in the air?

1.70816 s explanation The vat is fired off at an inclination of θ with the horizontal, so the vertical component of the initial velocity is vᵧ = v₀sinθ To determine how long the vat is in the air, consider when the vertical displacement is again zero: y= y₀+vᵧt-½gt² 0= vᵧt-½gt² 0= 2vᵧt-gt² 0= t(2vᵧ-gt) t= (2vᵧ)/g The trivial solution t=0 must be ignored

How far can Swimmer 2 cover in 30 seconds? LOOK AT FIGURE 3 ON GOOGLE DOC 1) 70 m 2) 50 m 3) 90 m 4) 60 m 5) 100 m 6) 40 m 7) 20 m 8) 80 m 9) 10 m 10) 30 m

10) 30 m explanation: In the first 30 s, Swimmer 2 has moved from 0 m to 30 m

189.6 km explanation: d=v₁t= (79 km/h) (2.4 h) = 189.6 km

A dog pulls on a pillow with a force of 7.4 N at an angle of 20.8° above the horizontal b) What is the y component of this force?

2.62779 N explanation Fᵧ= Fsinθ Fᵧ= (7.4 N) sin20.8° = 2.62779 N

A physics book is moved once around the perimeter of a table of dimensions 1 m by 3 m. What is the distance traveled? 1) 1 m 2) 0 m 3) 3 m 4) 5 m 5) 8 m 6) None of these 7) 2 m 8) 4 m 9) 6 m 10) 7 m

5) 8 m explanation: The distance traveled is the perimeter of the table: d= 2l + 2w = 2(1 m) + 2(3 m) = 8 m

SEE FIGURE 7 ON GOOGLE DOC A ball is thrown and follows the parabolic path shown. Air friction is negligible. Point Q is the highest point on the path. Points P and R are the same height above the ground. Describe an arrow of the acceleration of the ball at point R (if any).

The arrow would be pointing straight down explanation Since air friction is negligible, the only acceleration on the ball after being thrown is that due to gravity, which acts straight down

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