Phys 001

Units used in physics for mass, time and length

L - meter Mass - kilogram t - seconds

A car starts from rest and travels for 5 s w/ a uniform acceleration of 1.5 m/s^2. If the brakes are applied for 3.0 s a)how fast is the car going at the end of braking period? b) how far has the car gone?

a) V = Vo + at V = 0 + 1.5(5) V = 7.5 V =Vo + at V = 7.5 -2(3) = 1.5 m/s b)V^2 = Vo^2 + 2 a delta x (1.5)^2 = (7.5)^2 + 2(-2)delta x delta x = 13.5 V^2 = Vo^2 + 2 a delta x (7.5)^2 = (0)^2 + 2 (1.5) delta x delta x = 18.75 18.75 + 13.5 = 32.25 m

A person measures the hight of a building by walking out a distance of 46 m from its base and shining a light beam at the top. When the beam is elevated at an angle of 39 degrees with respect to the horizontal. a) if the flash light is held at a height 2.00 m. find the height of the building. b) calculate the length of the light beam

a) tan theta = y/x tan 39 degrees = y/ 46 = 37.3 + 2 m = 39.3 actual height b) r^2 = x^2 + y^2 r^2 = 46^2 + 37.3^2 r = 59.2

If a car is traveling at a speed of 28 m/s, is the diver exceeding the speed limits of 55 mi/h?

(28.0 meters/s) x (1 mi/1609 meters) = (1.74 x 10^-2 mi/s) x (60 s/1 min) x (60 min/1 hr) = 62.6 miles per hour

Area (L^2) equation How to find total area of a space

length x width Multiply each of the found areas to get the final total area

Acceleration units

m/s ^ 2 cm/s ^ 2 ft/s ^ 2

A baseball is hit so that it travels straight upwash after being struck by the bat. A fan observes that it takes 3 s for the ball to reach its maximum height. Find a) the balls initial velocity b) the height it reaches?

a) V = Vo + gt 0 = Vo + -9.8 (3) Vo = 29.4 b) delta x = Vot + 1/2 gt^2 delta x = 29.4 (3) + 1/2 (-9.8) ( 3)^2 delta x = 44.1

A certain car is capable of accelerating at a rate of .60 m/s^2 . How long does it take for this car to go from a speed of 55 mi/h to a speed of 69 mi/h?

convert from mi/h to m/s (55 mi/h) x (1609 m / 1 mi) x (I h/ 60 min) x (1 min / 60 sec) = 24.60 (60 mi/h) x ( 1609 m/ 1 mi) x ( 1h/ 60 min) x ( 1 min/ 60 sec) = 26.82 V = Vo + at 26.82 = 24.60 + .60t t = 3.7s

Two points are given in polar coordinates (r, theta)= (2 m, 50 degrees) and (5m , -50 degrees). What is the distance between them.

(2 m, 50 degrees) x = r cos theta 2 cos 50 = 1.29 y= r sin theta 2 sin 50 = 1.53 (1.29, 1.53) (5m , -50 degrees) x = r cos theta 5 cos - 50 = 3.21 y= r sin theta 5 sin -50 = -3.83 (3.21, -3.83) then apply distance formula d = square root (x2 - x1)^2 + (y2 - y1)^2

A jet plane has a takeoff speed of 75 m/s and can move along the runway at an average acceleration of 1.3 m/s^2. If the length of the runway is 2.5 km will the plane be able to use the runway safely.

V^2 = Vo^2 + 2a delta x 75^2 = 0^2 + 2 (1.3) delta x delta x = 2163.46 m convert back to km = 2.16 km

The driver of a 1.00 x 10 ^3 kg car traveling on the interstate at 35 m/s slams on his brakes to avoid hitting a second vehicle in front of him which had to come to rest because of congestion. After the braces are applied a constant kinetic friction force magnitude of 8.00 x 10^3 N acts on the car. a) what minimum distance should the brakes be applied to avoid a collision with the other vehicle? b) if the distance between the vehicles is initially only 30 m at what speed would the collision occur?

a) - f delta x = 1/2 mv final^2 - 1/2m v initial ^2 -8 x 10^3 delta x = 0 - 1/2(1000)(35)^2 delta x = 76.6 m b) - f delta x = 1/2 mv final^2 - 1/2m v initial ^2 -8 x 10^3 x 30 = 1/2 (1000)v final^2 - 1/2 (1000) (35 ^2 = 27.3

A driver of mass m drops from a board 10.0 m above the waters surface. Neglect air resistance a) use conservation of mechanical energy to find his speed 5.00 m above the waters surface b) find his speed as he hits the water?

a) 1/2 m v initial ^2 + mgyinitial = 1/2 m v final ^2 + mgy final v initial = 0 1/2 m 0^2 + m (9.8)(10) = 1/2 m v final ^2 + m(9.8)(5) 1/2 0^2 + (9.8)(10) = 1/2 v final ^2 + (9.8)(5) v final = 9.90 m/s b) find speed as he hits the water 1/2 m v initial ^2 + mgyinitial = 1/2 m v final ^2 + mgy final y final = 0 1/2 m 0 ^2 + m(9.8)10 = 1/2 m v final ^2 + m(9.8)0 (9.8)10 = 1/2 v final ^2 v final = 14 m/s

Two horses are pulling a barge with mass 2.00 x 10^ 3 kg along a canal. The cable connected to the first horse makes an angle of theta = 30 degrees with respect to the direction of the canal while the cable connected to the second horse makes theta2 = -45 degrees. Find the acceleration of the barge starting at rest, if each horse exerts a force of magnitude 6.00 x 10^12 N on the barge. Ignore forces of resistance on the barge.

a) F1 cos 30 = 600 N cos 30 = 5.20 x10 ^2 N F2 cos -45 600 cos -45 = 4.24 x 10^2 N 5.20 x10 ^2 N + = 4.24 x 10^2 N = 944 N b) F1 sin 30 = 600 N sin 30 = 300 N F2 sin -45 600 sin -45 = - 424 N 300 N + - 424 N = -124 N c) f = ma 944 = 2000 kg a = .472 m/s^2 -124 = 2000 kg a =.0620 m/s^2 d) tan = y/x tan = .0620 / .472

Express the location of the fly in polar coordinates... given (2,1)

r = square root 2^2 + 1^2 = r = 2.2 theta = inverse tan y/x = inverse tan 1/2 = 26.6 degrees. (2.2 , 26.6 degrees)

A fathom is a unit of length, usually reserved for measuring the depth of water. A fathom is 6 ft in length. The distance from earth to the moon is 250,000 miles. Use the approximation given to find the distance in fathoms

(250,000 mi/ 1) x (5280 ft / 1 mi) x ( 1/ 6 ft) = 2 x 10 ^8 fathoms

Vector Examples: A car travels 20.0 km due north and then 35.0 km in a direction 60.0 degrees west of north. Using a graph, find the magnitude and direction of a single vector that gives the net effect of the cars trip. This vector is called the cars resultant displacement. a)Find the horizontal and vertical components of the d = 1.00 x 10^2 m displacement of a superhero who flies from the top of a tall building at an angle of 30 degrees off the building downwards. b) suppose instead the superhero leaps in the other direction along a displacement vector B to the top of a flag pole where the displacement components are given by Bx = -25.0 m and By = 10.0m. Find the magnitude and direction of the displacement vector. A shopper leaves home and drives to a store located 7.00 km away in a direction 30 degrees North of east. Leaving the store the shopper drives 5.00 km in a direction 50.0 degrees west of north to a restaurant. Find the distance and direction from the shoppers home to the restaurant.

20 km + 35 km = 55 km Look at example from graph from section 1.9 Look at example from 1.10 a) Ax= A cos theta = 1x10^2 cos -30 degrees = 86.6m Ay = A sin theta = 1x10^2 sin -30 degrees = -50 m b) B = square root (Bx^2 + By^2) = square root (-25^2 + 10^2) = 26.9m theta = inverse tan (By/Bx) = tan inverse(10/-25) = -21.8 Take note wants to be in quadrant 2 so add 180 degrees Rx = A cos theta + B cos theta 7 cos 30 degrees + 5 cos (50 + 90degrees = 140) = 2.23 Ry = A sin theta + B sin theta 7 sin 30 degrees + 5 sin (50 + 90degrees = 140) = 6.71 R = Square root Rx^2 + Ry^2 = Square root (6.71^2 + 2.23^2) = 7.07 km To find theta = theta = inverse tan (Ry/Rx) = inverse tan(6.71/2.23) = 71.76 degrees.

One dimensional motion with constant acceleration examples: A race car starting from rest accelerates at a constant of 5.00 m/s^2. a) what is the velocity of the car after it has traveled 1.00 x10^2 ft? b) how much time has elapsedCalculate the average velocity two different ways? A typical jetliner lands at a speed of 1.60 x 10^2 mi/h and decelerates at the rate of 10 (mi/h)s. If the plane travels at a constant speed of 1.60 x 10^2 mi/h for 1.0 second after landing before applying the brakes, what is the total displacement of the aircraft between touchdown on the runway and coming to rest?

Convert 100 ft to meters = 30.5 meters a) Displacement vs Velocity V^2 = Vo^2 + 2a delta x square root V^2 = square root Vo^2 + 2a delta x = V = Vo^2 + 2a delta x = V = 0^2 + 2(5)(30.5) = 17.5 m/s b) V = Vo + at 17.5 = 0 + 5t = 3.5 sec Convert 1.60 x 10^2 mi/h to m/s (160mi/h)x(1609 m/1 mi)x(1 h/3,600 s) = 71.5 s Solve for initial displacement: delta x = Vot + 1/2 a t^2 delta x = 71.5 (1) + 1/2 (0)(1)^2 = 71.5 m Solve for final displacement: Convert declaration to m/s^2 -10 (mi/h)/s x (1609 m/ 1 mi) x (1 h/3600 s) = -4.47 m/s^2 V^2 = Vo^2 + 2a delta x 0^2 = 71.7^2 + 2(-4.47)delta x = 572 m 572 + 71.5 = 644 m

A ballistic pendulum is a device used to measure the speed of a fast moving projectile such as a bullet. The bullet is fired into a large block of wood suspended from some light wires. The bullet embeds in the block and the entire system swings up to a height of h. It is possible to obtain the initial speed if the bullet by measuring h and the two masses. As an example of the technique assume that the mass of the bullet m1 =5 g and the mass of m2 = 1 kg and h = 5 cm. a) find the velocity of the system after the bullet embeds in the block b) calculate the initial speed of the bullet.

a) 1/2 (M1 + M2) V^2 = (M1 + M2)gy All the masses will cancel out due to being equal then get rid of the 1/2 2 x (1/2 V^2) = (gy) x 2 V^2 = (gy) V^2 = 9.8 (0.5)x 2 V = square root 9.8 (0.5) x 2 V = 0.990 m/s b) initial speed of bullet momentum (M1 V1 + M2 V2) initial = (M1 V1 + M2 V2) final .005 kg V1 +M2(0) = (1.005 kg) (.990 m/s) .005 kg V1 = (1.005 kg) (.990 m/s) V1 = 199 m/s

Find the polar coordinates example: The cartesian coordinates of a point in the xy - plane are (x,y) = (-3.50 m, -2.50 m). Find the polar coordinates Convert (r, theta) = (5.00m, 37 degrees) to rectangular coordinates

Polar Coordinate (r, theta) To find r = square root r^2 = square root (-3.50^2) + (-2.50^2) = 4.30 theta = tan theta = y/x = -2.5/-3.5 = .714 = tan theta theta = tan inverse of ( .714) = 36 degrees Take note of which quadrant it originated in, the final answer has to be the same area. 36 degree + 180 = 216 degrees Final Answer = (4.30, 216 degrees) x coordinate = r cos theta = x y coordinate r sin theta = y

Two points in a rectangular coordinate system have the coordinated (5, 3) and (-3, 4) where the units are in cm. determine the distance between the points.

Use distance formula d = square root (x2 - x1)^2 + (y2 - y1)^2

A golf ball with mass 5.00 x 10^-2 is struck with a club. The force of the ball varies from zero when contact is made up to some maximum value and then back to zero when the ball leaves the club as the graph of force vs time. Assume that the ball leaves the club with a velocity of 44 m/s. a) find the magnitude of the impulse due to the collision b) estimate the duration of the collison and the average force acting on the ball?

Use impulse equation

Dimensional Analysis Problem: Show that the expression V = Vo + at is dimensionally correct, where V and Vo represent velocities, a is acceleration and t is a time interval. Find a relationship between an acceleration of constant magnitude (a), speed (v), and distance (r) from the origin for a particle traveling in a circle.

V = Vo + at at = (L/t^2)(t) = L/t L/t = L/t + L/t L/t = L/t a = L/t v = L/t t = L/v a = L/t^2 = (we don't want to use time) L/t^2 = L/(L/v)^2 = L/(L^2)/(v^2) =

The best leaper in the animal kingdom is the puma, which can jump to a height of 3.7 m when leaving the ground at an angle of 45 degrees. With what speed must the animal leave the ground to reach that height?

V^2 = Vo^2 - 2 g delta y 0^2 = Vo^2 - 2 (-9.8) 3.7 Vo = 8.52 m/s Vo sin (45) = 8.52 Vo = 12.04 m/s

Ch 4 ???? The hockey puck struck by a hockey stick is given an initial speed of 20 m/s on a frozen pond. The puck remains on the ice and slides 1.20 x 10^ 2 m, slowing down steadily until it comes to rest. Determine the coefficient of kinetic friction between the puck and ice.

Vo = 20 m/s delta x = 1.20 x 10^ 2 m

Killer whales are known to reach 32 ft in length and have a mass of over 8,000 kg. They are also very quick able to accelerate up to 30 mi/h in a matter of seconds. Dis regarding the considerable drag force of water, calculate the average power a killer whale with a mass of 8 x10^3 kg would need to generate to reach a speed of 12 m/s in 6 secs.

W = Delta KE = 1/2 m v final ^2 - 1/2 m v initial ^2 1/2 (8,000) (12)^2 - 0 = 57600 J 57600 J / 6 sec = 9.6 x 10^4 watts

A block with mass of 5.00 kg is attached to the horizontal spring with a spring constant of 4.00 x 10^2 N/m. The surface the block rests upon is frictionless. If the block is pulled out to Xi = .050 m and released a) find the speed of the block when it first reaches the equilibrium point. b) find the speed when x = .025 m c) repeat part a) if friction acts on the block with the coefficient Mk = .150

a) 1/2 m v initial ^2 + 1/2 K X initial ^2 = 1/2 m v final ^2 + 1/2 K X final ^2 at equlaibrim so 1/2 K X final ^2 = 0 which will cancel out 1/2 m 0^2 + 1/2 K X initial ^2 = 1/2 m v final ^2 1/2 K X initial ^2 = 1/2 m v final ^2 divide my m on both sides and get rid of the 1/2 (2) x 1/2 K X initial ^2/m = (2) x 1/2 m v final ^2/m K X initial ^2 / m = v final ^2 400 N / 5 kg (.05)^2 = .447 m/s b) 1/2 m v initial ^2 + 1/2 K X initial ^2 = 1/2 m v final ^2 + 1/2 K X final ^2 V initial = 0 1/2 m 0 ^2 + 1/2 K X initial ^2 = 1/2 m v final ^2 + 1/2 K X final ^2 1/2 K X initial ^2 - + 1/2 K X final ^2 = 1/2 m v final ^2 x2 (1/2 K X initial ^2 - + 1/2 K X final ^2) = (1/2 m v final ^2) x2 K X initial ^2 - K X final ^2 = m v final ^2 (K X initial ^2 - K X final ^2) /m = m v final ^2 /m (K X initial ^2 - K X final ^2) /m = v final ^2 K (X initial ^2 - X final ^2) / m = v final ^2 400 M (.05 ^2 - .025 ^2) / 5 kg = v final ^2 v final = .387

A rectangle has a length of 2.0 +- .2 and width 1.5+- .1. a) find the area, b) find the perimeter and uncertainty

a) 2 x 1.5 = 3 2.0 +- .2 = .2/2.0 =.1 = 10% 1.5 +- .1 >1/1.5 = .066 = 6.6% 10 + 6.6 = 16.6% 3 x 16.6 = 50%/100 = .5 3.0 +- .5 b) 2(a + b) 2(2.0 + 1.5) = 7 2(.167 + .167) = .6 7.0 +- .6

a) Find the gravitational force exerted by the sun on a 70 kg man located at the earths equator at noon. When the man is closest to the sun b) calculate the gravitational force of the sun on the man at midnight, when he he is farthest from the sun? c)calculate the difference in acceleration due to the sun between noon and midnight.

a) F = G(Mass person x Mass sun)/ r^2 r = 1.49 x 10 ^11 - 6.38 x 10^ 6 G = 6.67 x10^-11 F = 6.67 x10^-11(70 x 1.99 x 10^30)/ (1.49 x 10 ^11 - 6.38 x 10^ 6) ^2 = .41540 N b)F = G(Mass person x Mass sun)/ r^2 r = 1.49 x 10 ^11 + 6.38 x 10^ 6 F = 6.67 x10^-11(70 x 1.99 x 10^30)/ (1.49 x 10 ^11 + 6.38 x 10^ 6) ^2 = .415 N c) F = ma .415 - .41533 = 70 a a = 1 x10 ^-6 m/s^2

An air boat with a mass of 3.50 x 10^2 kg including the passenger has an engine that produces a net horizontal force of 7.70 x 10^2 N after for accounting of forces of resistance. a) find the acceleration of the the airboat b) starting from rest, how long does it take the airboat to reach a speed of 12 m/s? c) after reaching that speed the pilot turns off the engine and drifts to a stop over a distance of 50 m find the resistance force assuming its constant.

a) F = ma 7.70 x 10 ^2 N = 3.50 x 10 ^2 a a = 2.20 m/s^2 b) V =Vo + at 12 = 0 + 2.20 t t = 5.45 s c) V^2 = Vo^2 + 2 a delta x 0^2 = 12^2 + 2 a 50 a = -1.44 F= ma F = 3.50 x10^2 x -1.44 = F F = 504 N

A man of mass m = 75 kg and a woman m = 55 kg stand facing each other on an ice rink, both wearing skates. The woman pushes the man w/ a horizontal force of F = 85 N in the positive x -direction. Assume the ice is frictionless. a) what is the mans acceleration? b) what is the reaction force acting on the woman? c) calculate the woman's acceleration?

a) F = ma 85 = 75 kg a a = 1.12 b) reaction force on a woman -85 c) -85 = 55 a a = -1.55 m/s^2

A rocket moves straight upward starting from rest with an acceleration of 29. m/s^2. It runs out of fuel at the end of 4.00 s and continues to coast upward reaching a maximum height before falling back to earth. a) Find the rockets velocity and position at the end of 4.00 s b) find the max. height the rocket reaches c) find the velocity the instant before it hits the ground.

a) V = Vo + at V = 0 + 29.4(4) V = 118 m/s To find postion delta y = Vot + 1/2 a t^2 delta y = 0(4) + 1/2(29.4)(4)^2 delta y = 235 m b) Max. height V = Vo + gt 0 = 118 + -9.8 t t = 12 s delta y = Vot + 1/2 a t^2 Y final - Y finial = Vot + 1/2 a t^2 Y final - 235 = 118(12) + 1/2 (-9.8) 12^2 = 945 m c) 0 - 945 = 0t + 1/2 (-9.8) t^2 t = 13.9 s V = 0 - 9.8(13.9) = -136

A long jumper leaves the ground at an angle of 20 degrees to the horizontal and speed of 11 m/s a) how long does it take to reach the max. height? b) what is the maximum height? c) how far does he jump?

a) V = Vo sin theta - gt 0 = 11 sin 20 - 9.8 t t = .384 s b) delta y = Vo sin theta t + 1/2 gt^2 delta y = 11 sin 20 (.384) + 1/2 (-9.8).384^2 = .722 m c) .384 x 2 = .768 delta x = Vo cos theta t delta x = 11 cos 20 (.768)

A ball is thrown vertically upward with a speed of 25 m/s. a) how high does it rise? b) how long does it take to reach the highest point c) how long does the ball take to hit the ground after it reaches its highest point? d)what is the velocity when it returns to the level from which it started?

a) V^2 = Vo^2 + 2 g delta x 0^2 = 25^2 + 2 (-9.8) delta x delta x = 31.9 b) V = Vo + gt 0 = 25 + (-9.8) t t = 2.55 s c) ?? d) V = Vo + gt V = 0 + -9.8 (2.55) = -25m

Acessna air craft has a liftoff speed of 120 km/h a)what minimum constant acceleration does the air craft require if it is to be airborne after a takeoff run of 240 m? b) how long does it take the air craft?

a) Velocity vs Displacement Convert km/h to m/s (120 km/h) x (120000 m / 1 km) x (1h / 60 mins) = 33.3 V^2 = Vo^2 + 2a delta x 33.3^2 = 0^2 + 2 a 240 a = 2.3 b) V = Vo + at 33.3 = 0 + a2.3 a = 14.4

Free Falling Examples: A ball is thrown from the top of a building with an initial velocity of 20.0 m/s straight upward, at an initial height of 50.0 m above the ground. The ball just misses the edge of the roof units way down. Determine a) the time needed for the ball to reach its maximum height b) its maximum height c)the time needed for the ball to return to the height from which it was thrown and the velocity of the ball at that instant. d) the time needed for the ball to reach the ground, e) the velocity and position of the ball at t = 5.00 s. Neglect air drag

a) Velocity vs Time V = Vo + gt 0 = 20 m/s + 9.8(t) = 2.04 s b)Max. height --- Displacement vs Time delta y = Vot + 1/2 gt^2 delta y = 20 m/s(2.04) + 1/2(9.8)(2.04)^2 = 20.4 m c) Time needed to return --- Displacement vs Time delta y = Vot + 1/2 gt^2 0 = 20 m/s (t)+ 1/2(-9.8)(t)^2 = t = 4.08 s Velocity of the ball at that instant V = Vo + gt V = 20 + 9.8(4.08s) = -20 m/s d)time to reach the ground ---- displacement vs time delta y = Vot + 1/2 gt^2 -50 = 20 (t) + 1/2(-9.8)t^2 factor out a t = 5.83 s e) find the velocity and position V = Vo + gt V = 20 + - 9.8 (5) = -29 Position means displacement - delta y = Vot + 1/2 gt^2 delta y = 20(5) + 1/2(-9.8)(5)^2 = -22.5 m

A compact disc rotates from rest up to an angular velocity of -31.4 rad/s in a time of 0.892 s a) what is the angular acceleration of the disc, assuming the angular acceleration is uniform? b) through what angle does the disc turn while coming up to that speed? c) if the radius of the disc is 4.45 cm find the tangential velocity of a microbe riding on the rim of the disc when t = 0.829 s d) what is the magnitude of the tangential acceleration of the microbe at the given time?

a) W = Wo + alpha t -31.4 = 0 + alpha (0.892) alpha = -35.2 rad/s^2 b) delta theta = delta theta = Wot + 1/2 alpha t^2 delta theta = 0(.892) + 1/2 (-35.2) (.892)^2 delta theta = -14 rad c) Vt = rw 4.45 cm convert to m = .0445 m Vt = (.0445 m) (-31.4) = -1.40 m/s d) a = r alpha a = .0445 m (-35.2) = -1.57 m/s^2

A person travels by car from a city to another with different constant speeds between pairs of cities. Drives for 30 min at 80 km/h, 12 min at 100 km, and 45 min at 40 km/h and spends 15 min eating lunch and buying gas. a) determine the average speed for the trip b) determine the distance between the initial and final cities

a) add up all the times then divide by 60. b) convert each mins to hours and multiply by the km/h given with it. (30 min/1) x (1 hr/ 60 min) = .5 x 80 km/h = 40 do this for each set then add up the values for the total distance

A wheel rotates with a constant angular acceleration of 3.50 rad/s^2. If the angular velocity of the wheel is 2.00 rad/s^2 at t = 0 a) through what angle does the wheel rotate between t = 0 and t = 2 seconds? Give an answer in radians and in revolutions. b) what is the angular velocity of the wheel at t = 2 seconds c) what angular displacement in revolutions results while the angular velocity found in part b) doubles?

a) delta theta = Wot + 1/2 alpha t^2 delta theta = 2(2) + 1/2 (3.5) (2)^2 delta theta = 11 rad, revolutions = (11 rad / 1) ( 1 rev/ 2 pie rad) = 1.75 revs b) W= Wo + alpha t = 2 + 3.5(2) = 9 rads/ s c) 9 x 2 = 18 (doubled) W^2 = Wo^2 + 2 alpha delta theta 18^2 = 9^2 + 2(3.5)delta theta delta theta = 34.7 rad / 2 pie rad = 5.52 rev

A place kicker must kick a football from point 36 m from the goal. Half the crowd hopes the ball will clear the cross bar = 3.05 m high when kicked the ball leaves the ground with a speed of 20 m/s at an angle of 53 degrees to the horizontal. a)by how much does the ball need clear or fall short of clearing the cross bars? b) does the ball approach the cross bars while still rising or while falling?

a) delta x = Vot + 1/2 a t^2 36 = 20 cos 53 t + 1/2 0 t^2 t = 2.99 s delta x = Vot + 1/2 a t^2 delta x = 20 sin 53(2.99) +1/2(-9.8)(2.99)^2 = 3.94 3.94 - 3.05 = .89 b) V= Vo + gt V = 20 sin theta - 9.8(2.99) V = -13.3 falling

A boy throws a baseball onto a roof and it rolls back down and off the roof with a speed of 3.75 m/s. If the roof is pitched at 35 degrees below the horizon and the roof edge is 2.5 m above the ground find a) the time the baseball spends in the air b) the horizontal distance from the roof edge to the point where the baseball lands on the ground?

a) delta y = Vo sin theta t + 1/2 gt^2 2.5 = 3.75 sin 35 t - 4.9 t^2 t = .527 s b) delta x = Vo cos theta (.527) + 1/2 0 (.527)^2 delta x = 3.75 cos 35 + 1/2 0 (.527)^2 delta x = 1.62

A block has a mass of 4.00 kg resting on a slope that makes an angle of 30 degrees with the horizontal. If the coefficient of static friction between the block and surface rests upon .650 a) find the normal force b) the maximum static friction force c) the actual static friction force required to prevent the block from moving? d) will the block begin to move or remain at rest?

a) find normal force n = mg cos theta n = 4 kg(9.8) cos 30 n = 33.9 N b) Max. static fiction force Fs = Ms n Fs = .650 x 33.9 = 22.1 N c) Actual static friction force Fs = mg sin theta fs = 4(9.8) sin 30 = 19.6 N d) Will the block begin to move or remain at rest? max is 22.1 and actual is 19.6 so not enough force

A tennis player tosses a tennis ball straight up and then catches it after 2.00 s at the same height as the point of release. a) what is the acceleration of the ball while it is in a flight? b) what is the velocity of the ball when it reaches its maximum height ? c) the initial velocity of the ball and d) the maximum height it reaches?

a) gravity -9.8 b) 0 bc its not moving at the max height c) delta x = Vot + 1/2 gt^2 0 = Vo(2) + 1/2(-9.8)(2)^2 Vo = 9.8 d) V^2 = Vo^2 + 2 g delta x 0^2 = 9.8^2 + 2 (-9.8) delta x. delta x = 4.9m

Calculate the magnitude of the normal force on a 15 kg block in the following circumstances a) the block is siting on a level surface b) the block is resting on a surface tilted up at a 30 degrees angle with respect to the horizontal c) the block is resting on the floor of an elevator that is accelerating upwards 3.00 m/s^2 d) the block is on level surface and a force of 125 N is exerted on it at an angle of 30 degrees above the horizontal

a) n = mg 15 x 9.8 = 147 b) n = mg cos theta n = 147 cos 30 n = 127. 3 N c) n = ma + mg n = 15 x 3 + (147) = 192 N d) n = mg - Fapplied sin theta n = 147 - 125 sin 30 n = 84. 5 N

A horse is harnessed to a sled having the same mass of 236 kg including supplies. The horse must exert a force exceeding 1240 N at an angle of 35 degrees in order to get the sled moving. Treat the sled as a point particle. a) calculate the normal force on an sled when the magnitude of the applied force is 1240 N b) find the coefficient of static friction between the sled and the ground beneath it. c) find the static friction force when the horse is exerting a force of 6.20 x 10^2 N on the sled at the same angle.

a) n = mg - Fapplied sin theta n = 236(9.8) - 1240 sin 35 n = 1601.56 N b) Ms = F cos theta / N 1240 cos 35 / 1601.56 N = .634 c) Ms = 6.20 x 10^2 cos 35 = 507.87 N

For the triangle shown - HW 1 number 45 a) what is the length of the unknown side b)tangent of theta c) sin of theta

a) use a^2 + b^2= c^2 to find unknown side. b) and c) SOH CAH TOA

A man returning from a successful ice fishing trip pulls a sled loaded with salmon. The total mass of the sled and salmon is 50 kg and the man exerts a force of magnitude 1.20 x10^2 N on the sled by pulling on the rope. a) how much work does he do on the sled if the rope is horizontal to the ground theta = 0 degrees and pulls the sled 5 m? b) how much work does he do on the sled is theta = 30 degrees and he pulls the sled the same difference? Treat the sled as a point particle, so details such as the point of attachment of the rope make no difference. c) At a coordinate position of 12.4 m the man lets up on the applied force. A friction force of 45 N between the ice and the sled brings the sled to a rest at a coordinate position of 18.2 m. How much work does the friction do on the sled.

a) w = d x n w = 5 m x 120 N = 600 J b) w = F cos d w = 120 cos 30 (5) = 520 J c) -45(18.2 - 12.4) = -261 J

Displacement Examples: A turtle and a rabbit engage in a foot race over a distance of 4.00 km. The rabbit runs .500km and then stops for a 90.0 min nap. Upon awakening he remembers the race and runs twice as fast finishing the course in a total time of 1.75 hour, winning the race. a) calculate the average speed of a rabbit b) what was his average speed before he stopped for a nap? A train moves slowly along a straight portion of track according to the graph of position vs time. Find a)the average velocity for the the total trip b)the average velocity during the first 4.00 s of motion c) Average velocity during the next 4.00 s of motion d) the instantaneous velocity at t = 2.00 s and t = 9.00s. Look at graph on section 2.1 example 2. A baseball player moves in a straight line path in order to catch a fly ball hit to the outfield. Her velocity as a function of time is shown. Find her instanoeus acceleration at points A,B,C. look at graph in section 2.1 example 3.

a)Finding average speed Average speed = total distance / elapsed time 4.00km/1.75 hr = 2.29 km/h b)Convert mins to hours 90 mins = 1.5 hrs total time - nap time = 1.75 hr - 1.5 hr = .25 hr = total time before he stopped .25 = (d1 /V1) + (d2/2V2) = (.5/V1) + (4 - .5 = 3.5/2V2) Next, solve by making a common denominator, multiply 2 = = (2 x.5/2 x V1) + (3.5/2V2) Solve math to get V = 9 a)Delta x / delta t = 10 - 0/ 12 - 0 = .83 m/s b) 4-0/4-0 = 1 m/s c) 4-4/8-4 = 0 m/s d) 2-0/2-0 = 1 m/s e) 4.5 - 0/ 9 - 3 side note - 3 for letter e is from tangent line Point A = 2-0/1-0 = 2 m/s^2 Point B = 4-4/3-2 = 0 m/s^2 Point C = 2 - 4/ 4 - 3 = -2 m/s^2

HELP Relative Velocity Examples: An Alaskan rescue plane drops a package of emergency rations to standard hikers. The plane is traveling horizontally at 40 m/s at the height of 1.00 x 10^2 m above the ground. Neglect air resistance. a) Where does the package hit the ground relative to the point at which it was released? b) What are the horizontal and vertical components of the velocity of the package before it hits the ground? c) what is the angle of impact? A ball is thrown upward from the top of a building at an angle of 30 degrees above the horizontal and with an initial speed of 20 m/s. The point of release is 45 above the ground. a)How long does it take for the ball to hit the ground? b) Find the balls speed at impact c)find the horizontal range of the ball. Neglect air resistance. A jet plane traveling horizontally at 1.00 x 10^ 2 m/s drops a rocket from a considerable height. The rocket fires it engines accelerating at 20 m/s in the x-direction while falling under the influence of gravity in the y-direction. When the rocket has fallen 1.00 km a) find its velocity in the y-direction b) find velocity in x direction c) find the magnitude and direction of its velocity. Neglect air drag. The boat is heading north as it crosses a river with a velocity of 10 km/h. The river has a uniform velocity of 5 km/h due east. Determine the magnitude and direction of the boats velocity with respect to the observer on the river bank.

a)displacement vs time first find the time??? delta x = Vot + 1/2 at^2 delta x = 40(4.52) + 1/2(0)(4.52)^2 = 181 m b) horizontal component Vx = Vo cos theta Vx = 40 cos 0 degrees (bc going forward, if going backward then 180 degrees) = 40 m/s vertical component Vy = 40 sin 0 + gt ---why????? Vy = 40 sin 0 + (9.8)4.52 = -44.3 m/s c) tan theta (Vy/Vx) theta = tan inverse (-44.3/40.0) = -48 degrees a) how long for ball to hit the ground Vo y = Vo + sin theta Voy = 20 sin 30 = 10 Delta y = Vot + 1/2 g t^2 -45 = 10t + 1/2 9.8(t)^2 = 4.22 s b) find speed of impact V = Vo + gt V = 10 +(-9.8)(4.22) = -31.4 Vo cos theta = 20 cos 30 = 17.3 V = squareroot -31.4^2 + 17.3^2 = 35.9 c) finding horizontal range displacement = delta x = Vo(x) t delta x = 17.3 (4.22s) = 73.1 m a)displacement vs velocity In y- direction V^2 = Vo^2 + 2a delta x convert 1.00 km - to meters (1.00 km/1) x (1 x 10^3 m/1km)= 1000 m V^2 = 0 + 2(-9.8)(-1000) = -140 b) find velocity in x direction Find time V = Vo + gt -140 = 0 + -9.8t = t = 14.3 s V = Vo + at V = 100 + (20)(14.3) = 386 c) Find magnitude V = squareroot Vy^2 + Vx^2 V = Squareroot 386^2 + (-140)^2 = 411 Find direction theta = inverse tan (Vy/Vx) = -140/386 = -19.9 degrees a) find magnitude = squareroot 10^2 + 5^2 = 11.18 b) find direction theta = inverse tan 5/10 = 26.6 degrees