Q130 / 160 · TS EAMCET 2020 (14 Sep Shift 2) · Physics
A force acts on a $30 \mathrm{~g}$ particle in such a way that the position of the particle as a function of time is given by $x=\alpha t^2$, where $x$ is in metre, $t$ is in seconds and $\alpha=1 \mathrm{~m} / \mathrm{s}^2$. The work done during the first $4 \mathrm{~s}$ is
- Correct answerA0.96 J
- B0.45 J
- C0.49 J
- D0.53 J
Solution
Given, mass of particle, $m=30 \mathrm{~g}=3 \times 10^{-2} \mathrm{~kg}$ Displacement of the particle is given as
$x=\alpha t^2$
$\therefore$ Velocity, $v=\frac{d x}{d t}=\frac{d}{d t} \alpha t^2 \Rightarrow v=2 \alpha t$
Acceleration, $a=\frac{d v}{d t}=\frac{d}{d t}(2 \alpha t) \Rightarrow a=2 \alpha$
$\therefore$ Force on the particle,
$\begin{aligned}
F & =m a=3 \times 10^{-2} \times 2 \alpha & \\
& =6 \times 10^{-2} \mathrm{~N} & {[\because \alpha=1] }
\end{aligned}$
$\therefore$ Work done during first $4 \mathrm{~s}$,
$\begin{array}{rlrl}
W & =\int_0^4 F d x=\int_0^4 6 \times 10^{-2} d x & \\
& =6 \times 10^{-2} \int_0^4 d x=6 \times 10^{-2} \int_0^4 v d t \quad[\because d x=v d t] \\
& =6 \times 10^{-2} \int_0^4 2 \alpha t d t=6 \times 10^{-2} \times 2 \alpha \cdot\left[\frac{t^2}{2}\right]_0^4 \\
& =12 \times 10^{-2} \times 8 & {[\because \alpha=1]} \\
& =0.96 \mathrm{~J}
\end{array}$
Hence, the correct answer is option: 'A'.