Physics

Physics

Problem

A 0.5 kg0.5\,\text{kg} stone is accidentally released from the top of a 20 m20\,\text{m} tall building. It passes a window ledge that is 12 m12\,\text{m} above the ground and is caught instantaneously by a worker standing there. The worker then throws the stone straight back up so that it just reaches the original point of release. a. (i) Taking the ground as the zero level, calculate the change in gravitational potential energy of the stone as it falls from the top of the building to the ledge. b. (ii) How much work is done by the force of gravity on the stone during this fall? c. (iii) Determine the work the worker must do on the stone to return it from the ledge to the top of the building. d. (iv) After the stone completes the entire cycle described above, what is the net change in its gravitational potential energy with respect to the ground? State a reason for your answer.
Intuitive explanation

The key insight is that you're really solving two separate motion problems connected by energy: first, find the stone's speed when it reaches the window using kinematics, then use that speed as your starting point for the upward throw. For the return trip, "just reaches" means the stone arrives with zero velocity—so you can either use v2=v02−2gΔhv^2 = v_0^2 - 2g\Delta h or recognize that the worker must throw it with exactly the speed it had when caught, since the heights are identical.

Intuitive explanation

The key insight is that you're really solving two separate motion problems connected by energy: first, find the stone's speed when it reaches the window using kinematics, then use that speed as your starting point for the upward throw. For the return trip, "just reaches" means the stone arrives with zero velocity—so you can either use v2=v02−2gΔhv^2 = v_0^2 - 2g\Delta h or recognize that the worker must throw it with exactly the speed it had when caught, since the heights are identical.

Browse the problem library