Question

A body of mass $5$ kg has momentum of $10$ kg m/s. When a force of $0.2$N is applied on it for $10$ seconds, what is the change in its kinetic energy?

• A. $1.1$J
• B. $2.2$J
• C. $3.3$J
• D. $4.4$J

Answer 1

Answer: (D)

$4.4$J

%Increase in momentum$\left(P\right)=n.$

Then,% Increase in kinetic energy$\left(E\right)=n^2+\frac{200n}{100}.$

\quad Initial Momentum $=P=10N-s$

Mass$=5kg$, Force $=0.2N.$

Time for which force acted $=10s.$

Now, Initial Kinetic Energy$=E=\frac{1}{2}m{v}^2$

$=\frac{1}{2}\left(m\right)(\frac{p}{m}{)}^2$

$=\frac{1}{2}\left(5\right)(\frac{10}{5}{)}^2$

$=10J$

Now, $\Delta P=F\times t{P}^{\prime }-P=2P^{\prime }$

$=P+2.$

we see that increase in momentum is $\left(\frac{2}{10}\right)\times 100=20$

Increase in Kinetic Energy$=44$

$=\frac{44}{10}$

Increase in Kinetic Energy $=4.4J.$