Work Power and Energy 5 Question 13
16. The displacement $x$ of a particle moving in one dimension, under the action of a constant force is related to the time $t$ by the equation $t=\sqrt{x}+3$ where $x$ is in metre and $t$ in second. Find
(1980)
(a) the displacement of the particle when its velocity is zero, and
(b) the work done by the force in the first $6 s$.
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Answer:
Correct Answer: 16. (a) zero (b) zero
Solution:
- Given, $t=\sqrt{x}+3$ or $\sqrt{x}=(t-3)\cdots(i)$
$\therefore \quad x=(t-3)^{2}=t^{2}-6 t+9\cdots(ii)$
Differentiating this equation with respect to time, we get
$$ \text { velocity } v=\frac{d x}{d t}=2 t-6 \cdots(iii) $$
(a) $v=0$ when $2 t-6=0$ or $t=3 s$
Substituting in Eq. (i), we get
$$ \begin{aligned} \sqrt{x} & =0 & or\\ x & =0 \end{aligned} $$
i.e. displacement of particle when velocity is zero is also zero.
(b) From Eq. (iii), speed of particle at $t=0$ is
$v _i=|v|=6 m / s$
and at $t=6 s$
$$ v _f=|v|=6 m / s $$
From work energy theorem,
Work done $=$ change in kinetic energy
$$ =\frac{1}{2} m\left[v _f^{2}-v _i^{2}\right]=\frac{1}{2} m\left[(6)^{2}-(6)^{2}\right]=0 $$
