A stone of mass 20g is released from a catapult whose rubber is stretched through 5cm. If the force constant of the rubber is 200Nm\(^{-1}\), calculate the speed with which the stone leaves the catapult.
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Correct Answer: Option
Explanation:
½Ke\(^{2}\) = ½mv\(^{6}\)
½ x 200 x (0.05)\(^{2}\) = ½ x 0.02 x v\(^{2}\)
v\(^{2}\) = 25
v = 5 m s\(^{-1}\)
½Ke\(^{2}\) = ½mv\(^{6}\)
½ x 200 x (0.05)\(^{2}\) = ½ x 0.02 x v\(^{2}\)
v\(^{2}\) = 25
v = 5 m s\(^{-1}\)