Q1 REVIEW

Quiz 1 Question 6 If the train accelerates at a -4 m/s^2 and the train has a mass of 6000kg. Find the net force acting on the train.

-24,000N

Quiz 1 Question 3 Emily drove to san Diego from Mesa, a distance of 300miles, at a speed of 60mph. She returned at a speed of 50mph. What was her total displacement?

0miles

Quiz 1 Question 5 Emily drove to san Diego from Mesa, a distance of 300miles, at a speed of 60mph. She returned at a speed of 50mph. What was her average velocity?

0mph

Quiz 1 Question 17 A bot rolls his toy car off the kitchen table at a speed of 4m/s. The table is 90cm tall. How far away from the base of the table does the car land?

107m

Forces Homework Question 7 A 65kg person dives into the water from a 10m platform. What is her speed as she enters the water? (round to the nearest tenth) ____m/s

14.1 m/s a= 650N/65kg Known variables: v_o= 0, x=10m, a= 9.8m/s^2 kinematic equation: v_f^2 = v_o^2 + 2ax v_f^2 = 0^2 +2(9.8m/s^2)(10m) square root both sides 14 m/s =14.1 m/s^2 (using 10m/s^2 as acceleration of gravity)

Quiz 1 Question 13 If an object accelerates from rest at a constant rate of 6m/s^2, how fast would it be travelling at the end of 3 seconds?

18 m/s

Forces Homework Question 3 A student, standing on a scale in an elevator at rest, sees that his weight is 840N. As the elevator rises, his weight increases to 1050N, then returns to normal. Determine the acceleration at the beginning of the trip. _____m/s^2

2.5 m/s^2 at rest Weight = 840N 840N/9.8m/s^2 = 85.7kg the reading of the scale is the normal force of the body, not actually the weight. F_net=ma, so F_N - W = 1050-840 = 210N a=F/m 210N/8507kg = 2.45m/s^2 2.5m/s^2 What would the acceleration be at the end of the trip? When the elevator slows to a stop at the 10th floor, his weight drops to 588N F=ma, so 588-840N = -252N -252N/8507kg = 2.94m/s^2

Quiz 1 Question 16 A model rocket fires horizontally off the edge of a cliff at a velocity of 50m/s. If the canyon is 100m deep, how far from the edge of the cliff does the model rocket land?

224m

Forces Homework Question 8 The same woman as in the previous question comes to a stop 2m below the surface of the water. What net force did the water exert on the swimmer? _____N

3,250N Find acceleration first Known variables: v_o= 14m/s, v_f= 0, x= 2m use kinematic equation v_f^2 = v_o^2 + 2ax 0 = 14^2 + 2*a*2 subtract 14^2 both sides 196 = 4a a = 49m/s^2 F=ma F_net= 65kg*49m/s^2 3185N = 3250N (using 10m/s^2 as acceleration of gravity)

Quiz 1 Question 7 A 700kg Nissan can go from rest to a speed of 45 m/s in 9.0s. What average net force acts on the car?

3,500 N

Quiz 1 Question 15 Starting from rest, a freely-falling object will fall in 10 seconds, a distance of about...

500m

Quiz 1 Question 4 Emily drove to san Diego from Mesa, a distance of 300miles, at a speed of 60mph. She returned at a speed of 50mph. What was her average speed?

54.5mph

Forces Homework Question 1 A 4600 kg helicopter accelerates upward at 2.0m/s^2. What lift force is exerted by the air on the propellers?

54280N or 55200N L = Lift, W = Weight F=L - W = ma L = ma + W = ma +mg = m(a+g) so, L = m(a+g) L = 4600kg*(2.0m/s^2+9.8m/s^2) =54280N =55200N (using 10m/s^2 as acceleration of gravity)

Forces Homework Question 4 A sign in an elevator states that the maximum occupancy is 20 persons. Suppose that the safety engineers assume the mass of the average rider is 75kg. The elevator itself has a mass of 500kg. The cable supporting the elevator can tolerate a maximum force of 30,000N. What is the greatest acceleration the elevator's motor can produce without snapping the cable? _____m/s^2

5m/s^2 Given values N=20, m_p=75kg, m_e=500, F_tol Total mass given by M_tot =N*m_p + m_e M_tot = 20*75kg+500 M_tot =2000kg f=ma to a=f/m ∑F_y = M_tot*a rearrange for acceleration a= (F_tot- F_T)/M_tot a= (30000N-(2000*9.8))/2000 = 5.2m/s^2 =5m/s^2 (using 10m/s^2 as acceleration of gravity) In summary, (maximum cable force) - Weight = (Total Mass)*acceleration

Forces Homework Question 5 A racecar has a mass of 710kg. It starts from rest and travels 40.0m in 3.0s. The car is uniformly accelerated during the entire time. What net force is acting on the car? (round to the nearest whole Newton) _____N

6,311N Kinematic formula describing an object traveling uniformly with a constant acceleration: (from x=v_0*t+1/2at^2) x = 1/2at^2 40.0m=1/2a(3.0s)^2 a = 8.89m/s^2 F=ma F_net= 710kg*8.89m/s^2 = 6311.9N 6300 (If rounded to sig figs)

Quiz 1 Question 2 Emily drove to san Diego from Mesa, a distance of 300miles, at a speed of 60mph. She returned at a speed of 50mph. What was her total distance?

600miles

Forces Homework Question 6 Suppose that a 1000kg car is traveling at 25m/s (about 55mph). Its brakes can apply a force of 5000N. What is the minimum distance required for the car to stop? _____m

62.5m F=ma to a=F/m a = -5000/1000kg = -5m/s^2 Known variables: a = -5m/s^2, v_f = 0(b/c car is stopping), and v_0 = 25m/s use kinematic equation: V_f^2 = V_i^2 + 2a(delta x) Solve for x x= (v_f^2 - v_o^2)/2a x=(0^2 - 25m/s^2) / 2*-5m/s^2 x = 62.5m

Quiz 1 Question 1 Suppose you are driving at 30m/s for 6 seconds. At time 2s, you look at the radio to change the station and do not return your eyes to the road until 5s. How much distance did your car travel while you were looking at the radio?

90m

Quiz 1 Question 11 Describe what a graph of position vs. time graph for an object moving with constant velocity looks like.

sloping line

There are more questions in

the blue physics notes I took.

Kinematic Equations

Use to solve problems in one dimension with uniform acceleration (free-fall). Equations which arise from the definition of velocity. They relate the position of the cg to velocity components. Cg being centre of gravity. *(No x) V_f = V_i + at t = (v_f-v_i)/a *(No v_f) delta x = v_i*t + 1/2at^2 *(No t) V_f^2 = V_i^2 + 2a(x_f - x_i) V_f^2 = V_i^2 + 2a(delta x) *(No a) delta x = 1/2(v_o+v_f)t *(No v_i) delta x = v_f*t - 1/2at^2 x= displacement, v_f= final velocity, v_i = initial velocity, t= time, a= acceleration V_i is sometimes shown as V_o *How to solve? Define the unknown and the three knowns. usually, a = 9.8m/s^2 acceleration of gravity. v_i = 0 if the object was originally at rest and the third known variable depends on what is given in the problem Find the right equation and solve for the unknown.

Freefall motion Question 1 An object falls freely from rest on earth. a) Find how far it falls in 3 seconds. b) Find the time for it to reach a speed of 25m/s. c) Find the time required for it to reach 300m.

a) delta x = v_i*t + 1/2at^2 (no t) y=0*3s + (1/2)(9.8m/s^2)(3s^2) y=44.1m b) V_f = V_i + at 25m/s = 0m/s +9.8m/s^2*t 25/9.8 = t=2.6s c) delta x = v_i*t + 1/2at^2 300m = 0*t + (1/2)(9.8m/s^2)t^2 Square root out both sides. √300/(0.5*9.8) +t t = 3.5s

Freefall motion Question 5 A body is thrown downward with an initial speed of 20m/s on Earth. What is the: a) acceleration of the object b) displacement after 4s c) time required to reach a speed of 30m/s

a) he is accelerating dues to gravity at 9.8m/s^2 b) x = v_i*t + 1/2at^2 x = (-20.0m/s)(4.0s) + (1/2)(-9.8m/s^2)(4.0s^2) = c) v_f = v_i + at t = (v_f-v_i)/a t = (-30m/s - (-20.0m/s)) / 9/8m/s^2 = t = 1.02s

Freefall motion Question 3 A ball is dropped from rest height of 80m above ground a) What is the speed just as it hits the ground? b)How long does it take for it to reach the ground?

a) v_final^2 = v_intial^2 + 2ax v_f^2 = 0^2 + 2(9.8m/s^2)(80m) = v_f = 39.6 m/s b) x = v_i*t + (1/2)at^2 80m=0*t + (1/2)(9.8m/s^2)(t^2) t = 4.04s

Freefall motion Question 6 A student throws his worthless lab partner off a 120m high bridge with an initial downward speed of 10m/s. a) How long does it take the deadbeat to hit the ground below? b) How fast is he going a the moment of impact?

a) x = v_i*t + 1/2at^2 -120 = (-10m/s)*t + (1/2)(10m/s^2)(t^2) (gravity made to 10m/s for easier calculations) then factoring the quadratic formula -120 = -10t - 5t^2 t^2+2t-24 = 0 (t+6)(t-4) = 0 t= -6 or t=4 Since there is no such thing as negative time... t = 4s b) v_f = v_i + at -10m/s + -9.8m/s^2 (4s) = v_f = -49.2m/s

Freefall motion Question 4 A marble dropped from a bridge strikes the water in 6.0 seconds a)Find the speed with which it strikes the water b) Find the height of the bridge.

a)v_f = v_initial +at v_f = 0+ (9.8m/s^2)(6.0s) = v_f = 58.8 m/s b)x = V_i*t + 1/2at^2 x= 0*t + 1/2(9.8m/s^2)(6s^2) = x = 176.4 m

Quiz 1 Question 9 On a velocity vs. time graph, the slope represents....

acceleration

Quiz 1 Question 10 What does the area under a velocity vs. time graph represent?

displacement

Freefall motion Question 2 On the moon, gravity is 1.67m/s^2. Objects fall at a slower rate than on Earth. How far does a freely falling object on the moon fall in 3 seconds?

displacement = V_initial t + 1/2at^2 d=0*3s + (.5)(1.67m/s^2)(3^2) displacement = 7.5m

Quiz 1 Question 14 The distance a freely falling object travels each surrounding second would be

greater than the previous second

Forces Homework Question 2 The maximum force that a grocery bag can withstand without ripping is 250 N. Suppose that the bag is filled with 20 kg of groceries and lifted with an acceleration of 5.0 m/s^2. Do the groceries stay in the bag?

no F=ma F=20kg*5.0m/s^2 =100N The 100N force is less than the maximum force of 250N, so the bag will hold.

Quiz 1 Question 12 Describe what a graph of position vs. time graph for an object moving with constant acceleration looks like.

parabola

Quiz 1 Question 8 On a position vs. time graph, the slope represents....

velocity