Laws Of Motion Question 129

Question: A 15 kg block is initially moving along a smooth horizontal surface with a speed of $ v=4m/s $ to the left. It is acted by a force $ F $ , which varies in the manner shown. Determine the velocity of the block at $ t=15 $ seconds. Given that, $ F=40\cos ( \frac{\pi }{10} )t $

Options:

A) 12.5 m/s

B) 8.5 m/s

C) 20 m/s

D) 9.5 m/s

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Answer:

Correct Answer: A

Solution:

[a] Change in liner momentum $ \Delta P=\int{Fdt} $

$ 15(v _{f}+u)=\int\limits _{0}^{15}{40\cos ( \frac{\pi }{10} )tdt} $

$ v _{f}=-4+\frac{40}{15}{{[ \frac{\sin (\pi /10)t}{\pi /10} ]}^{15}} _{0}=-4+\frac{400}{15\pi }(-1) $

$ =-12.5m/s $



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