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A body of mass $4 \mathrm{~kg}$ is accelerated upon by a constant force, travels a distance of $5 \mathrm{~m}$ in the first second and a distance of $2 \mathrm{~m}$ in the third second. The force acting on the body is
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Verified Answer
The correct answer is:
$6 \mathrm{~N}$
Distance travelled by the body in $\mathrm{n}^{\text {th }}$ second is given by
$$
\begin{aligned}
\mathrm{S}_{\mathrm{n}} &=\mathrm{u}+\frac{\mathrm{a}}{2}(2 \mathrm{n}-1) \\
5 &=\mathrm{u}+\frac{\mathrm{a}}{2}(2 \times 1-1) \\
5 &=\mathrm{u}+\frac{\mathrm{a}}{2} \\
2 &=\mathrm{u}+\frac{\mathrm{a}}{2}(2 \times 3-1) \\
2 &=\mathrm{u}+\frac{5}{2} \mathrm{a}
\end{aligned}
$$
Solving Eqs. (i) and (ii), we get
$$
a=-\frac{6}{4} \mathrm{~m} / \mathrm{s}^{2}
$$
body is decelerating
$$
\text { mass }=4 \mathrm{~kg}
$$
and
$$
\mathrm{F}=\mathrm{m} \times \mathrm{a}=4 \times \frac{6}{4}=6 \mathrm{~N}
$$
$$
\begin{aligned}
\mathrm{S}_{\mathrm{n}} &=\mathrm{u}+\frac{\mathrm{a}}{2}(2 \mathrm{n}-1) \\
5 &=\mathrm{u}+\frac{\mathrm{a}}{2}(2 \times 1-1) \\
5 &=\mathrm{u}+\frac{\mathrm{a}}{2} \\
2 &=\mathrm{u}+\frac{\mathrm{a}}{2}(2 \times 3-1) \\
2 &=\mathrm{u}+\frac{5}{2} \mathrm{a}
\end{aligned}
$$
Solving Eqs. (i) and (ii), we get
$$
a=-\frac{6}{4} \mathrm{~m} / \mathrm{s}^{2}
$$
body is decelerating
$$
\text { mass }=4 \mathrm{~kg}
$$
and
$$
\mathrm{F}=\mathrm{m} \times \mathrm{a}=4 \times \frac{6}{4}=6 \mathrm{~N}
$$
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