Work Energy and Power - Result Question 29
32. Consider a car moving along a straight horizantal road with a speed of $72 km / h$. If the coefficient of static friction between road and tyres is 0.5 , the shortest distance in which the car can be stopped is
[1994]
(a) $30 m$
(b) $40 m$
(c) $72 m$
(d) $20 m$
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Answer:
Correct Answer: 32. (b)
Solution:
- (b) Force due to friction $=$ kinetic energy
$\mu m g s=\frac{1}{2} m v^{2}$
$[.$ Here, $v=72 km / h=\frac{72000}{60 \times 60}=20 m / s$ ] or , $s=\frac{v^{2}}{2 \mu g}=\frac{20 \times 20}{2 \times 0.5 \times 10}=40 m$
If a body of mass $m$ moves with velocity $u$ on a rough surface and stops after travelling a distance $S$ due to friction. Then, frictional force, $F=ma=$ $\mu R \Rightarrow ma=\mu mg \Rightarrow a=\mu g$
Using $v^{2}=u^{2}-2 a s$
$\Rightarrow 0=u^{2}-2 \mu g S$
$\Rightarrow S=\frac{u^{2}}{2 \mu g}$
