A cricket ball of mass 0.15 kg is thrown vertically up by a bowling machine so that it rises to a maximum height of 20 m after leaving the machine. If the part pushing the ball applies a constant force F on the ball and moves horizontally a distance of 0.2 m while launching the ball, the value of F (in N) is (g = 10 ms –2 ) ____.

Solution

Physics sir · 4 steps

The ball's 30 J at the top all came from the machine's push, so set F × 0.2 equal to it.

1. Energy needed. To rise 20 m, the ball must leave the machine with kinetic energy equal to the potential energy it gains at the top: mgh = 0.15 × 10 × 20 = 30 J.
2. Where it comes from. The ball starts at rest. Only the constant force F gives it this energy, and it acts over the 0.2 m stroke. By the work-energy theorem, the work done by F equals the kinetic energy at launch.

   F × d = mgh

3. Put in numbers. F × 0.2 = 30, so F = 30/0.2 = 150 N. As a check, the launch speed is v = √2gh = 20 m s⁻¹, and 1/2 × 0.15 × 20² = 30 J.
4. So. The force applied by the machine is 150 N.

The idea to keep. When a force launches a body over a short stroke, link the two ends with energy: work done by the force equals the kinetic energy given, which you can find from the height or speed it reaches.

Answer from the NTA final answer key. Solution by PrepWiser.