JAMB - Physics (2000 - No. 6)
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
Explanation
Given: mass of stone = 500g = 0.5kg, e = 20cm = 0.2m, velocity of stone on leaving the catapult = 40m/s
The potential energy stored in the catapult when extended = the kinetic energy (KE) of the stone
\(\frac{1}{2}\)Fe = \(\frac{1}{2}\)mv\(^2\)
F x 0.2 = 0.5 x 40 x 40
F = \(\frac{800}{0.2}\) = 4000 N = 4 x 10\(^3\) N
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