AP Physics 1- 3.2 Work - Exam Style questions - FRQs- New Syllabus

(a)
On Planet \( X \), the radius is greater while the planet’s mass is the same as Earth’s, so the gravitational field strength is smaller.

Because the sphere is released from the same angle \( \theta \), it travels through the same vertical distance from \( A \) to \( B \) in both situations. The work done by gravity depends on the gravitational force and the vertical displacement.

Since the downward gravitational force on the sphere is smaller on Planet \( X \), but the vertical distance traveled is the same, the work done by gravity is smaller. Therefore,

\( \boxed{W_X < W_E} \)

Equivalently, the decrease in gravitational potential energy from \( A \) to \( B \) is smaller on Planet \( X \) because \( g \) is smaller there.

(b)
The period of a simple pendulum is given by

\( T = 2\pi \sqrt{\dfrac{L}{g}} \)

On Planet \( Y \), the planet’s radius is the same as Earth’s but its mass is larger, so the gravitational field strength \( g \) is larger. If the string length stayed the same, the larger value of \( g \) would make the period smaller, because \( g \) is in the denominator. That would make \( T_Y < T_E \).

However, this new string is slightly elastic, so its length can change when the tension changes. On Planet \( Y \), the sphere has greater weight, so the tension in the string can be larger. That larger tension could stretch the string and increase its length \( L \). Since the period increases with \( L \), a large enough increase in string length could outweigh the effect of the larger \( g \), making the period longer instead. In that case, \( T_Y > T_E \).

So \( T_Y \) could be smaller than \( T_E \) because of the larger gravitational field strength, or it could be larger than \( T_E \) if the increased tension stretches the elastic string enough to increase the pendulum length significantly.