An object of mass $20 \mathrm{~kg}$. is displaced by $x=5 t^2 \mathrm{~m}$ (here $t$ is time) by the…
An object of mass $20 \mathrm{~kg}$. is displaced by $x=5 t^2 \mathrm{~m}$ (here $t$ is time) by the application of a force. Then the ratio of the work done in times $3 \mathrm{~s}$ and $5 \mathrm{~s}$ is
$2 / 3$
$4 / 9$
$3 / 5$
$9 / 25$
Solution
( Given, displacement of block, $x=5 t^2$
As, velocity $v=\frac{d x}{d t}$
Velocity of block, $v=\frac{d}{d t}\left(5 t^2\right)$
$
=5 \times 2 t=10 t \mathrm{~m} / \mathrm{s}
$
Now, velocity of block at $t=5 \mathrm{~s}$,
$
v_2=10 \times 5=50 \mathrm{~m} / \mathrm{s}
$
and velocity of block at $t=3 \mathrm{~s}$ is
$
v_1=10 \times 3=30 \mathrm{~m} / \mathrm{s}
$
Also, velocity of block at $t=0 \mathrm{~s}$,
$
v_0=10 \times 0=0 \mathrm{~m} / \mathrm{s}
$
Now, by work- K.E theorem, work done by a force $=$ Change in kinetic energy
So work done by force in first $3 \mathrm{sec}$,
$
W_1=\frac{1}{2} m v_1^2-\frac{1}{2} m v_0^2=\frac{1}{2} m\left(v_1^2-v_0^2\right)
$
And work done by force in first $5 \mathrm{sec}$,
$
W_2=\frac{1}{2} m v_2^2-\frac{1}{2} m v_0^2=\frac{1}{2} m\left(v_2^2-v_0^2\right)
$
Hence, ratio $\frac{W_1}{W_2}=\frac{\frac{1}{2} m\left(v_1^2-v_0^2\right)}{\frac{1}{2} m\left(v_2^2-v_0^2\right)}$.
$\begin{aligned} & =\frac{v_1^2-v_0^2}{v_2^2-v_0^2} \\ \text { or } \quad \frac{W_1}{W_2} & =\frac{v_1^2}{v_2^2} \quad\left(\text { As }, v_0=0\right) \\ \Rightarrow \quad \frac{W_1}{W_2} & =\left(\frac{30}{50}\right)^2=\frac{9}{25}\end{aligned}$