Physics 11

NewtonÕs Second Law Calculations

 

1.     A towrope is used to pull a 1750 kg car giving it an acceleration of 1.35 m/s2. What force does the rope exert? 2360 N

2.     A racing car undergoes a uniform acceleration of 4.00 m/s2. If the net force causing the acceleration is 3.00 x 103 N what is the mass of the car? 750 kg

3.     A 5.2 kg bowling ball is accelerated from rest to a velocity of 12.0 m/s as the bowler covers 5.0 m of approach before releasing the ball. What force is exerted on the ball during this time?  a = 14.4 m/s2; F = 74.9 N

4.     A high jumper, falling at 4.0 m/s, lands on a foam pit and comes to rest, compressing the pit 0.400 m. If the pit is able to exert an average force of 1200 N on the high jumper in breaking the fall, what is the jumper's mass? a = -20. m/s2; m = 60. kg

5.     On Planet X, a 50.0 kg barbell can be lifted by exerting a force of only 180 N.

a) What is the acceleration of gravity on Planet X? 3.6 m/s2

b) If the same barbell is lifted on Earth what minimal force is needed? F = 490 N

6.     A proton has a mass of 1.672 x 10-27 kg. What is its weight? F = 1.639 x 10-26 N

7.     A force of 20.0 N accelerates a 9.00 kg wagon at 2.0 m/s2 along the sidewalk.

a) How large is the frictional force? 2.0 N

b) What is the coefficient of friction? 0.023

8.     A 2.00 kg brick has a sliding coefficient (kinetic coefficient) of friction of 0.38. What force must be applied to the brick for it to move at a constant velocity? Fnet = 0 N; Fapp = 7.45 N (just enough to overcome friction)

9.     In bench-pressing 100. kg a weight lifter applies a force of 1040 N. How large is the upward acceleration of the weights during the lift? Fnet = Fapp Ð Fg therefore Fnet = 60. N and a = 0.60 m/s2

10.  An elevator that weighs 3.0 x 103 N (mass is 306 kg) is accelerated upward at 1.0 m/s2. What force does the cable exert to give it this acceleration? Fnet = FT Ð Fg therefore FT = ma + Fg = 3.31 x 103 N

11.  A person weighing 490 N (mass is 50. kg) stands on a scale in an elevator.

a) What does the scale read when the elevator is at rest? 490 N (Fg = FN the normal force is actually measured by the scale!)

b) What is the reading on the scale when the elevator rises at a constant velocity? 490 N (a = 0 m/s2)

c) The elevator slows down at Ð2.20 m/s2 as it reaches the desired floor. What does the scale read? From experience, you may know that when an elevator goes downward we feel a little bit of weightlessness. Here Fnet = FN - Fg by defining up as positive and down as negative. Since Fnet is Ð110 N (mass of 50. kg times Ð2.20 m/s2), we can rearrange the above to find that the normal force is 380 N and the person weighs less. You could do this by subtracting the g Ð a (9.80 Ð 2.20 m/s2) and then multiplying by the mass.

d) The elevator descends accelerating at Ð2.70 m/s2. What does the scale read? Same as above but with a larger downward acceleration so FN = 355 N.

e) What does the scale read when the elevator descends at a constant velocity? 490 N (a = 0 m/s2)

f) Suppose the cable snapped and the elevator fell freely. What would the scale read? Here Fnet = FN - Fg by defining up as positive and down as negative. Since Fnet is Ð490 N (mass of 50. kg times Ð9.80 m/s2), we can rearrange the above to find that the normal force is 0 N and the person weighs nothing.

g) What does the scale read if the elevator was to accelerate upwards at 1.0 m/s2? From experience, you may know that when an elevator goes upwards we feel a little bit of extra force downward on us. Here Fnet = FN - Fg by defining up as positive and down as negative. Since Fnet is +50. N (mass of 50. kg times +1.0 m/s2), we can rearrange the above to find that the normal force is 540 N and the person weighs more. You could do this by adding the g + a (9.80 + 1.0 m/s2) and then multiplying by the mass.

 

12.  When a 20.0 kg child steps off a 3.0 kg stationary skateboard with an acceleration of 0.50 m/s2, with what acceleration will the skateboard travel in the opposite direction? The force that child exerts on the skateboard is 10. N, so there is an equal but opposite force of 10. N on the skateboard. It is pushed away at 3.3 m/s2.

13.  Given the following diagrams determine the Fnet and the acceleration of the system. Then find the tension in each cable that connects the masses. Consider the pulleys and tabletops to be frictionless.

a)                                                                     b)

                                  

 

Fnet on system = Fg on 2.5 Ð FT + FT Ð Fg on 1.0       

F net on system = Fg of 1.0 kg Ð FT + FT

Fnet = Fg on 2.5 Ð Fg on 1.0

Fnet = Fg og 1.0 kg

msystema = m2.5g Ð m1.0g

msystema = mg

(3.5 kg)a = (2.5 kg)(9.80 m/s2) Ð (1.0 kg)(9.80 m/s2)

(3.5 kg)a = (1.0 kg)(9.80 m/s2)

a = 4.2 m/s2

a = 2.8 m/s2