Q170 / 180 · COMEDK 2018 · Physics
A body of mass $4 \mathrm{~kg}$ is accelerated upon by a constant force, travels a distance of $5 \mathrm{~m}$ in the 1st second and a distance of $2 \mathrm{~m}$ in the $3 \mathrm{rd}$ secon(d)The force acting on the body is
- A$2 \mathrm{~N}$
- B$4 \mathrm{~N}$
- Correct answerC$6 \mathrm{~N}$
- D$8 \mathrm{~N}$
Solution
We know that, displacement travelled by a moving body with constant acceleration a in $n$th second,
$$
s_{n}=u+\frac{1}{2} a(2 n-1)
$$
For first second, $n=1$
$$
\begin{aligned}
\therefore \quad s_{1} &=u+\frac{1}{2} a(2 \times 1-1) \\
5 &=u+\frac{a}{2} \quad \ldots(\mathrm{i})\left[\because s_{1}=5 \mathrm{~m}\right]
\end{aligned}
$$
For third second, $n=3$
$$
\begin{array}{rlr}
\therefore \quad s_{3} & =u+\frac{a}{2}(2 \times 3-1) \\
2 & =u+\frac{5 a}{2} & \ldots\left(\text { ii) }\left[\because s_{3}=2 \mathrm{~m}\right]\right.
\end{array}
$$
Subtracting Eq. (i) from Eq. (ii), we get
$$
\begin{aligned}
&-3=2 a \\
&\Rightarrow \quad a=\frac{-3}{2} \mathrm{~m} / \mathrm{s}^{2}
\end{aligned}
$$
Force acting on the body,
$$
F=m a=4 \times\left(\frac{-3}{2}\right)=-6 \mathrm{~N}
$$
Hence, force acting on the body in opposite direction of its motion will be $6 \mathrm{~N}$.
Hence, the correct answer is option: 'C'.