Q128 / 160 · TS EAMCET 2022 (19 Jul Shift 1) · Physics
A ball of mass $1 \mathrm{~kg}$ moves in a straight line with velocity $v=c x^a$, where $c=1$ (SI unit) and $\alpha$ is a constant. If the work done by the net force during its displacement from $\mathrm{x}=0$ to $\mathrm{x}=4 \mathrm{~m}$ is $128 \mathrm{~J}$, then the $\alpha$ is
- A1
- B$\frac{3}{2}$
- Correct answerC2
- D$\frac{1}{2}$
Solution
From work - Energy theorem
Work done $=$ Change in kinetic energy
$$
\begin{aligned}
128 & =\frac{1}{2} m v_2^2-\frac{1}{2} m v_1^2 \\
128 & =\frac{1}{2} v_2^2-\frac{1}{2} v_1^2
\end{aligned}
$$
...(i) $(\because \mathrm{vm}=118)$
We have $\mathrm{v}=\mathrm{cx}^\alpha$
$$
\begin{aligned}
& \mathrm{v}_1 \text { at } \mathrm{x}=0=0 \\
& \mathrm{v}_2 \text { at } \mathrm{x}=4=(4)^\alpha=(2)^{2 \alpha}
\end{aligned}
$$
From equation (i)
$$
\begin{aligned}
& 128=\frac{1}{2} \times\left[(2)^{2 \alpha}\right]^2 \\
& 2 \times 128=(2)^{4 \alpha} \\
& (2)^8=(2)^{4 \alpha} \\
& \Rightarrow 4^\alpha=8 \Rightarrow \alpha=2
\end{aligned}
$$
Hence, the correct answer is option: 'C'.