G L Mittal and TARUN MITTAL Solutions for Chapter: Friction, Exercise 2: NUMERICALS

A car is moving on a straight road at a speed of $20 {\mathrm{ms}}^{-1}$. The coefficient of friction between the tyres of the car and the road is $0.4$. Find the shortest distance within which the car can be stopped, $\left(g=10 {\mathrm{ms}}^{-1}\right)$.

 A car going at a speed of $7 \mathrm{m}/\mathrm{s}$ can be stopped by applying brakes in the shortest distance of $10 \mathrm{m}$. Show that the total frictional force opposing the motion, when brakes are applied, is $\frac{1}{4}th$ of the weight of the car. $\left(\mathrm{g}=9.8 {\mathrm{ms}}^{-2}\right)$

A fireman of mass $80 \mathrm{kg}$ slides down a pole during a rescue mission. The force of friction is constant at  $720 \mathrm{N}$. Find the acceleration of the man. $\left(g=10 \mathrm{m}/{\mathrm{s}}^2\right)$

A block placed on a rough inclined plane of length $4.0 \mathrm{m}$ just begins to slide down when the upper end of the plane is $1.0 \mathrm{m}$ high from the ground. Calculate the coefficient of static friction.

A block rests on a plane of variable inclination. Find the inclination of the plane with the horizontal at which the block just slides down. The coefficient of friction between the block and the plane is $\mu =\frac{1}{\sqrt{3}}$.

A block of mass $0.05 \mathrm{kg}$ when placed on a rough $15°$ inclined plane slides down without acceleration. The inclination is then increased to $30°$. What would be the acceleration of the block ? $g=9.8 \mathrm{m}/{\mathrm{s}}^2$

A body is in limiting equilibrium on a rough plane inclined at $30°$ with the horizontal. When the inclination is increased to $45°$, the body slides down with acceleration. Find this acceleration. Take $g=9.8 \mathrm{m}/{\mathrm{s}}^2$.

A body slides down from rest from the top of a $6.4 \mathrm{m}$ long rough plane inclined at $30°$ with the horizontal. Find the time taken by the block in reaching the bottom of the plane. Take ${\mu }_k=0.2$ and $g=9.8 \mathrm{m}/{\mathrm{s}}^2$