A catapult used to hold a stone of mass 500g is extended by 20cm with an applied force F. If the stone leaves with a velocity of 40m/s, the value of F is
- 4.0 x 102N
-
2.0 x 103N
- 4.0 x 103N
- 4.0 x 104N
The correct answer is: B
Explanation
To find the applied force F on the catapult, we can use the concepts of work and energy. The work done on the stone by the applied force is equal to the kinetic energy gained by the stone when it leaves the catapult.
Given: mass of the stone, m = 500g = 0.5kg,
extension of the catapult, d = 20cm = 0.2m,
velocity of the stone when it leaves, v = 40m/s
The kinetic energy gained by the stone when it leaves the catapult is given by: K .E = \(\frac{1}{2}mv^2\) = \(\frac{1}{2} \times 0.5 \times40^2\) = 400J
The work done by the applied force is given by: W = F X d = F x 0.2
Since the work done on the stone is equal to the kinetic energy gained:
F x 0.2 = 400
F = \(\frac{400}{0.2}\) = 2000N = 2.0 x 10\(^3\)N
The value of the applied force F is 2.0 x 10\(^3\)N
There is an explanation video available .