Question

A body of mass $2kg$ initially at rest moves under the action of an applied horizontal force of $7N$ on a table with coefficient of friction $=\mathrm{0.1.}$ The change in kinetic energy in $10s$ is: $\left(g=10m/s^2\right)$

• A. 425 J
• B. 625 J
• C. 725 J
• D. 825 J

Answer 1

Answer: (B)

$a=\frac{\sum F}{m}=\frac{F_{ex}-\mu mg}{m}$
$=\frac{7N-(0.1)(2kg)\left(10m/s^2\right)}{2kg}$
$=\frac{5N}{2}=2.5m{s}^{-2}$
After $t=10s$
$v=u+at=0+(2.5)(10)=25m{s}^{-1}$
$\Delta K=\frac{1}{2}m{v}^2-\frac{1}{2}m{u}^2=\frac{1}{2}$
(2) $(25{)}^{2}J=625J$.