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A body of mass $4 \mathrm{~kg}$ is moving with momentum of $8 \mathrm{~kg} \mathrm{~ms}^{-1}$. A force of $0.2 \mathrm{~N}$ acts on it in the direction of motion of the body for 10 seconds. The increase in kinetic energy in joules is
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Verified Answer
The correct answer is:
$4.5$
$$
\begin{aligned}
m & =4 \mathrm{~kg}, p=8 \mathrm{~kg} \mathrm{~m} / \mathrm{s} \\
F & =0.2 \mathrm{~N}, t=10 \mathrm{sec}
\end{aligned}
$$
From Newton's IInd law
Force $=$ rate of change of momentum
$$
\begin{aligned}
& F=\frac{p_2-p_1}{t} \\
& 0.2=\frac{m v-8}{10} \\
& 2=4 \times v-8 \\
& 4 v=10 \\
& \Rightarrow \quad v=\frac{5}{2} \mathrm{~m} / \mathrm{s} \\
& \therefore \quad \text { Final momentum } p_2=m v=4 \times \frac{5}{2} \\
& =10 \mathrm{~kg} \mathrm{~m} / \mathrm{s} \\
&
\end{aligned}
$$
\begin{aligned}
m & =4 \mathrm{~kg}, p=8 \mathrm{~kg} \mathrm{~m} / \mathrm{s} \\
F & =0.2 \mathrm{~N}, t=10 \mathrm{sec}
\end{aligned}
$$
From Newton's IInd law
Force $=$ rate of change of momentum
$$
\begin{aligned}
& F=\frac{p_2-p_1}{t} \\
& 0.2=\frac{m v-8}{10} \\
& 2=4 \times v-8 \\
& 4 v=10 \\
& \Rightarrow \quad v=\frac{5}{2} \mathrm{~m} / \mathrm{s} \\
& \therefore \quad \text { Final momentum } p_2=m v=4 \times \frac{5}{2} \\
& =10 \mathrm{~kg} \mathrm{~m} / \mathrm{s} \\
&
\end{aligned}
$$
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