Work Power and Energy 5 Question 13

(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:

  1. 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

$$ =\frac{1}{2} m\left[v _f^{2}-v _i^{2}\right]=\frac{1}{2} m\left[(6)^{2}-(6)^{2}\right]=0 $$



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