Question

$A 15$ kg block is initially moving along a smooth horizontal surface with a speed of $v = 4$ m/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 theat $F=40$ cos $(\frac{\pi}{10})t$

• A. $12.5$ m/s
• B. $8.5$ m/s
• C. $20$ m/s
• D. $9.5$ m/s

Answer 1

Answer: (A)

Change in linear momentum $\Delta p=\int Fdt$
$15\left(v_{f}+u\right)=\int \limits_{0}^{15} 40cos \left(\frac{\pi}{10}\right)t dt$
$v_{f}=-4+\frac{40}{25}\left[\frac{sin \left(\pi/10\right)t}{\pi/10}\right]_{_0}^{^{15}} =-4$
$4+\frac{400}{15 \pi}\left(-1\right)=-12.5 \, m\ s$