Question

A body of mass 5 kg is moving with a momentum of 10 kg m/s. A force of 0.2 N acts on it in the direction of motion of the body for 10 sec. Find the increase in its kinetic energy.

• A. 2.8J
• B. 3.2J
• C. 3.8J
• D. 4.4J

Answer 1

The correct option is D 4.4J

$t=10$sec

$10=5\times v[P=mv]$

$v=2m/s$

As $F=0.2N$

$a=\frac{0.2}{5}m/s^2=\frac{1}{25}m/s^2[a=\frac{F}{m}]$

at $t=0,KE=\frac{1}{2}\times s\times \left(2{)}^2\right[KE=\frac{1}{2}m{v}^2]$

$K{E}_i=10J$

Now, at $t=10$sec, $v_f=u+at$

$v_f=g+\frac{1}{25}\times 10=2.4m/s^2$

$\therefore K{E}_f=\frac{1}{2}\times 5\times (2.4{)}^2$

$=14.4J$

$\therefore △KE=14.4-10$

$=4.4J$