, etkina, gentile, van heuvelen s laws ap: assignment 4

College Physics, Etkina, Gentile, Van Heuvelen

Chapter 3: Applying Newton’s Laws

Copyright 2014 © Pearson Education, Inc.

AP: Assignment 4 Solutions

Questions:

1. (c) As the car accelerates along the road, the static friction force is exerted by the road in the

direction of motion and the friction force is exerted by the air in the opposite direction.

3. (b) Pulling is easier. Pulling the mower at an upward angle will reduce the normal force and

friction that opposes its motion.

9. (a) The acceleration of the modified Atwood machine will be less than g, since there is

tension in the string pulling the other mass.

12. (a) The force diagrams are shown below, for the case where you exert a force FYon TV

,

pushing the TV box to the right, with a toaster oven box next to it.

For the TV box, other forces involved are: (1) FE on TV

, the force exerted by Earth; (2) FTO on TV

,

the force by the toaster oven box; and (3) Ns on TV

, upward normal force by surface.

Similarly, for the toaster oven box, we have (1) FE on TO

, the force exerted by Earth; (2) FTV on TO

,

the force by TV; and (3) Ns on TO

, upward normal force on the toaster box by surface.

(b) We show below the force diagrams for the case where you exert a force FYon TO

, pushing the

toaster oven box to the right, with a TV oven box next to it.




13. When the Atwood machine is in motion, one block always moves up while the other moves

down. The directions of acceleration are different even though the magnitudes are the same.

Since acceleration is a vector quantity, strictly speaking the two masses do not have the same

accelerations.

Problems:

5. (a) With

Fx,F

y( ) = -200 N,100 N( ), the

magnitude of the force vector is

F = F

x

2 + Fy

2 = -200 N( )2

+ 100 N( )2

= 223.6 N

The direction is

We choose the latter since the vector is in the

second quadrant.

(b) With

Fx,F

y( ) = 300 N,400 N( ), the magnitude

of the force vector is

F = F

x

2 + Fy

2 = 300 N( )2

+ 400 N( )2

= 500 N

The direction is

We

choose the former since the vector is in the first

quadrant.




(c) With the magnitude

of the force vector is

F = F

x

2 + Fy

2 = 400 N( )2

+ -300 N( )2

= 500 N

The direction is

or 37° below the +x-axis.

6. Since the mass is in equilibrium, the magnitude of the vector sum of the forces exerted by

ropes 2 and 3 must be equal to the force exerted by rope 1, which is 400 N.

7. Applying Newton’s second law to both x- and y-directions, we obtain

With T

1 on K= 400 N, we find

T

2 on K=

T1 on K

sin53°=

400 N

sin53°= 500 N

which gives T

3 on K= T

2 on Kcos53° = 500 N( )cos53° = 300 N .




10. (a) The force diagram for the block is shown to

the right. The forces involved are: (1) FE on B

, the force

exerted by Earth, and (2) NS on B

, the upward normal

force by the surface.

(b) The force diagram for the sled is shown to the right.

The forces involved are: (1) FE on sled

, the force exerted

by Earth, (2) TR on sled

, the force the rope, (3) NS on sled

,

upward normal force by the surface, and (4) fS on sled

,

friction force by surface.

(c) The force diagram for the sled on the incline is

shown to the right. The forces involved are: (1)

FE on sled, the force exerted by Earth, (2)

TR on sled, the

force by the rope, (3) NS on sled

, upward normal force

by the surface, and (4) fS on Y

, friction force by surface

opposing the motion.

(d) The force diagram for the skydiver is shown to the

right. The forces involved are: (1) FE on D

, the force

exerted by Earth, and (2) FAir on D

, upward drag force by

air that balances the gravitational force, so that the

skydiver descends at a constant speed.




24. By ignoring friction, we find the acceleration of the car downhill to be

a = g sinq = 9.8 m/s2( )sin4.8° = 0.82 m/s2

The time required to complete the race is given by

d =1

2a Dt( )

2

Þ Dt =2d

a=

2 301 m( )0.82 m/s2

= 27.1s

and the speed of the car at the end of the race is

v2 = v

0

2 + 2ad Þ v = 2ad = 2 0.82 m/s2( ) 301 m( ) = 22.2 m/s

41. Sketch and translate A sketch of the situation is shown below.

We choose the car as the system, with Earth as the object of reference.

The coordinate system consists of a horizontal x-axis pointing in the

direction of the velocity and a vertical y-axis pointing upward.

Simplify and diagram Assume that the car can be modeled as a point-like

object. (b) The forces involved are: (1) FE on C

, the force exerted on the

car by Earth, (2) NS on C

, upward normal force by the surface, and (3)

fS on C, friction force by surface to stop the car. We ignore air resistance

here. The force diagram is shown to the right.

Represent mathematically The magnitude of the force of friction is given by fS on C

= mNS on C

, where s

is the effective coefficient of friction. Applying Newton’s second law to the x and y axes, we obtain




- fS on C

= max

NS on C

- mg = 0

respectively.

Solve and evaluate The equation can be solved to give the acceleration of the car: a

x=- m

sg. The

distance the car travels before coming to a stop is given by

v2 = v

0

2 + 2ad = v0

2 - 2msgd

With v

0= 60 mph = 26.8 m/s,we find the coefficient of static friction to be

In the above, we assume that the wheels roll to a stop without being locked, so the friction is static in nature. On the other hand, if the brake is applied too hard such that the wheels become locked, then the car will skid along the road, and the stopping force will be due to kinetic friction.

71. (a) The frictional force exerted by road surface on the tires is what propels the minivan to move

forward.

(b) Let D be the resistive force that opposes the motion. The force required to accelerate the minivan

is

F = D + ma = 300 N + 1560 kg( ) 2.0 m/s2( ) = 3420 N

, etkina, gentile, van heuvelen s laws ap: assignment 4

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