Question

A ball of mass 2 kg is thrown up with a velocity of 20 m s^–1.

1. Identify the sign of the work done by gravity on the ball during its upward motion and its downward motion.
2. If the ball reaches a height of 19.4 m, how much work was done by air resistance (assume g = 10 m s^–2).

Answer 1

Given,

Mass (m) = 2 kg

Initial velocity (u) = 20 m s^–1

Acceleration due to gravity (g) = 10 m s^–2

(i) When moving uphill, the displacement is upward, and the force of gravity works downward (opposite directions). As a result, gravity's work is negative. Both the displacement and the force of gravity are downward (in the same direction) during downward motion. Therefore, gravity's work is positive.

(ii) Initial kinetic energy of the ball,

`K = 1/2 "mu"^2 = 1/2 xx 2 xx (20)^2 = 400  "J"`

At the maximum height (19.4 m), the ball comes to rest, so its kinetic energy = 0.

Change in kinetic energy = Final KE – Initial KE = 0 – 400 = –400 J

Work done by gravity during the rise = –mgh = –(2 × 10 × 19.4) = –388 J

Using the work-energy theorem,

Total work done = Change in kinetic energy

Work done by gravity + Work done by air resistance = Change in kinetic energy

(–388) + W_air = –400

W_air = –400 + 388 = –12 J

Hence, the work done by air resistance = –12 J.