Work Energy And Power Ques 9
9. A block of mass $10$ $ kg$, moving in $x$ direction with a constant speed of $10 $ $ms^{-1}$, is subject to a retarding force $F=0.1 x $ $J / m$ during its travel from $x=20 $ $m$ to $30$ $ m$. Its final $KE$ will be:
[2015]
(a) $450 $ $J$
(b) $275 $ $J$
(c) $250 $ $J$
(d) $475 $ $J$
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Answer:
Correct Answer: 9.(d)
Solution:
- (d) Given, $m=10 $ $kg$
$u=10$ $ m / sec$
Retarding force, $F=0.1 x $ $J / m$
Acceleration, $a=\frac{F}{m}$
$a=\frac{0.1 x}{10}=0.01 x$
From, $v^{2}=u^{2}-2 a s$
$\Rightarrow v^{2}=(10)^{2}-2 \times 0.01 \int _{20}^{30} x d x$
$\Rightarrow v^{2}=100-0.02[\frac{x^{2}}{2}] _{20}^{30}$
$\Rightarrow v^{2}=100-\frac{0.02}{2}[(30)^{2}-(20)^{2}]$
$\Rightarrow v^{2}=100-0.01[900-400]$
$\Rightarrow v^{2}=100-0.01[500]$
$\Rightarrow v^{2}=95$ $ m / sec$
Final K.E. $=\frac{1}{2} m v^{2}$
$=\frac{1}{2} \times 10 \times 95 \Rightarrow 475 $ $J$
From, $F=m a$
$a=\frac{F}{m}=\frac{0.1 x}{10}=0.01 x=V \frac{d V}{d x}$
So, $ \int _{v_1}^{v_2} v d V=\int _{20}^{30} \frac{x}{100} d x $
$ \begin{aligned} &-\frac{V^{2}}{2}| _{V_1} ^{V_2}=.\frac{x^{2}}{200}| _{20} ^{30}=\frac{30 \times 30}{200}-\frac{20 \times 20}{200} \\ &=4.5-2=2.5 \\ & \frac{1}{2} m(V_2^{2}-V_1^{2})=10 \times 2.5 J=-25 J \end{aligned} $
Final K.E.
$ \begin{aligned} & =\frac{1}{2} m v_2^{2}=\frac{1}{2} m v_1^{2}-25=\frac{1}{2} \times 10 \times 10 \times 10-25 \\ & =500-25=475 J \end{aligned} $
BETA
