Question

A crate is pushed horizontally with $100N$ across a $5m$ floor. If the frictional force between the crate and the floor is $40N,$ then the kinetic energy gained by the crate is

• A. $200J$
• B. $240J$
• C. $250J$
• D. $300J$
• E. $500J$

Answer 1

Answer: (D)

$300J$

We know that,
$F-t=ma$
So, $100-40=ma$
$a=\frac{600}{m}$
Also, $v=\sqrt{2as}$
Kinetic energy gained by the crate
$KE=\frac{1}{2}m{v}^2$
$KE=\frac{1}{2}m\left(2as\right)$
$KE=\frac{1}{2}\times m\times (2\times \frac{60}{m}\times 5)$
$(\because s=5m)$
$KE=300J$