Home / AP Physics 1: 4.3 Conservation of Energy, the Work- Energy Principle, and Power – Exam Style questions with Answer- MCQ

AP Physics 1: 4.3 Conservation of Energy, the Work- Energy Principle, and Power – Exam Style questions with Answer- MCQ

Question

A rubber ball with mass \(0.20 \mathrm{~kg}\) is dropped vertically from a height of \(1.5 \mathrm{~m}\) above a floor. The ball bounces off of the floor, and during the bounce \(0.60 \mathrm{~J}\) of energy is dissipated. What is the maximum height of the ball after the bounce?

(A) \(0.30 \mathrm{~m}\)
(B) \(0.90 \mathrm{~m}\)
(C) \(1.2 \mathrm{~m}\)
(D) \(1.5 \mathrm{~m}\)

▶️Answer/Explanation

Ans:C

Question

A sled slides down a hill with friction between the sled and hill but negligible air resistance. Which of the following must be correct about the resulting change in energy of the sled-Earth system?

(A) The sum of the kinetic energy and the gravitational potential energy changes by an amount equal to the energy dissipated by friction.
(B) The gravitational potential energy decreases and the kinetic energy is constant.
(C) The decrease in the gravitational potential energy is equal to the increase in kinetic energy.
(D) The gravitational potential energy and the kinetic energy must both decrease.

▶️Answer/Explanation

Ans:A

Question

What is the instantaneous power due to the gravitational force acting on a 3-kilogram projectile the instant the projectile is traveling with a velocity of 10 meters per second at an angle of 30 degrees above the horizontal?
(A) 300 W
(B) 150 W
(C) − 150 W
(D) − 300 W

▶️Answer/Explanation

Ans:(C)

Power = work/time = ( fd cos 120°)/T = ( mg cos 120°)(D/T)
                   = (3)(10)( − 0.5)(10 m/s) = − 150 W

Question

A hockey puck of an unknown mass is sliding along ice that can be considered frictionless with a velocity of 10 meters per second. The puck then crosses over onto a rough floor that has a coefficient of kinetic friction equal to 0.2. How far will the puck travel before friction stops it?
(A) 2.5 m
(B) 5 m
(C) 25 m
(D) Depends on the mass

▶️Answer/Explanation

Ans:(C)  \(W=\Delta KE\)

       \(\mu ND\cos180^{\circ}=0-\frac{1}{2}mv^{2}\)

        \((\mu mgD)(-1)=-\frac{1}{2}mv^{2}\)

         \(D=v^{2}/2\mu g=100/(2\times0.2\times10)=25m\)

Question

 A dart is placed onto a spring. The spring is stretched a distance x . By what factor must the spring’s elongation be changed so that the maximum kinetic energy given to the dart is doubled?
(A) 1/2
(B) 2
(C) 4
(D) 2

▶️Answer/Explanation

Ans:(D) Conservation of energy:

                \(KE=EPE=\frac{1}{2}kx^{2}\)

                 Since we need \(x^{2}\) to double, increase \(x\) by \((2)^\frac{1}{2}=\sqrt{2}\).

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