Laws Of Motion Question 129
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:
Change in linear momentum $ \Delta P=\int{Fdt} $
$ 15(v _{f}+u)=\int\limits _{0}^{15}{40\cos ( \frac{\pi }{10} t )dt} $
$ v _{f}=-4+\frac{40}{15}{{[ \frac{\sin (\pi /10)t}{\pi /10} ]}^{15}} _{0}=-4-\frac{400}{15\pi }$
$ = -12.5, \text{m/s}$
BETA
