Physics: An Introduction Chapter 5 -- Selected Solutions

Chapter 5: Newton's Laws of Motion
Selected Solutions

Selected Solutions

Solutions to selected end of chapter problems from the Walker text book.

9. (a) According to Newton's 2^nd law F_av = ma_av with . Therefore,

Therefore, the average force is 5.15 kN opposite to the direction of motion.

(b) Since we know the average acceleration from part (a) above, we can treat the motion as if the acceleration is constant having the value of a_av. From the equations for motion with constant acceleration we can write

.

15. (a) By Newton's 3^rd law (of action and reaction) the parent and the child apply equal and opposite forces to each other. Therefore, the force experienced by the child is the same as the force experienced by the parent.

(b) The acceleration of the child is more than the acceleration of the parent. The child, who has less mass than the parent, must have a larger acceleration to keep the forces equal.

25. (a) The net force on the skier results in the skier's motion. This motion is parallel to the slope downhill, thus the direction of the net force is also parallel to the slope downhill. To get the magnitude we multiply the skier's mass times the parallel component of acceleration

(b) As the slope becomes steeper, the net force increases. As the incline increases, sinθ approaches 1. In this case, the skier is falling. As the incline decreases, sinθ approaches zero and the skier is standing still. In this case, the force of gravity is counteracted by the force of the ground on the skier.

35. (a) The direction of acceleration is upward. An upward acceleration results in an apparent weight greater than the actual weight because the force that the scale applies must exceed the weight to accelerate you upward.

(b) The magnitude of the acceleration can be determined by applying Newton's 2^nd law tot he y-direction

39. (a) The free-body diagram for the child is shown on the right.

The normal force is found by applying Newton's 2^nd law for the child in the vertical direction noting that a = 0.

(b) The free-body diagram for the chair is shown on the right.

The normal force is found by applying Newton's 2^nd law for the chair in the vertical direction noting that a = 0.

39. (a) The free-body diagram for the child is shown on the right.
The normal force is found by applying Newton's 2^nd law for the child in the vertical direction noting that a = 0.

(b) The free-body diagram for the chair is shown on the right.
The normal force is found by applying Newton's 2^nd law for the chair in the vertical direction noting that a = 0.