Question

Two blocks A and B having negligible separation between them are given velocity $4\,m\,{\sec ^{ - 1}}$ and $8\,m\,{\sec ^{ - 1}}$ respectively. Coefficient of friction between A and the surface is $0.1$ what should be the coefficient of friction between B and ground so that collision never takes place between A and B?

Answer 1

In order to solve this question we need to understand friction. Friction is stopping pull force which always act opposite in direction of motion of object. It is an effect of Newton's third's law which states that every action has an equal and opposite reaction. Friction is directly proportional to surface normal and hence coefficient of friction is proportionality constant. There are two types of coefficient of friction, first is kinetic and second is static. Kinetic friction acts on a moving object whereas static friction acts on a stationary body.

Complete step by step answer:
Let the two blocks move in the same direction.Velocity of block A is, ${v_A} = 4m{\sec ^{ - 1}}$ and coefficient of friction of A is, ${\mu _A} = 0.1$.Velocity of block B is, ${v_B} = 8m{\sec ^{ - 1}}$ and let coefficient of friction of B be ${\mu _B}$.
So acceleration of block A is, ${a_A} = - {\mu _A}g$
And acceleration of block B is, ${a_B} = - {\mu _B}g$
Here $g$ is acceleration due to gravity, $g = 10m{\sec ^{ - 2}}$
So using equation of motion for both blocks,
Distance covered by block A in time t is, ${s_A} = {u_A}t + \dfrac{1}{2}{a_A}{t^2}$
Putting values we get, ${s_A} = {v_A}t - \dfrac{1}{2}{\mu _A}g{t^2}$
${s_A} = 4t - \dfrac{1}{2}{t^2}$

Similarly Distance covered by block B in time t is, ${s_B} = {u_B}t + \dfrac{1}{2}{a_B}{t^2}$
Putting values we get, ${s_B} = {v_B}t - \dfrac{1}{2}{\mu _B}g{t^2}$
${s_B} = 8t - 5{\mu _B}{t^2}$
For two blocks not collide $({s_B} - {s_A}) > 0$
Putting values we get, $(4t - \dfrac{1}{2}{t^2} - 8t + 5{\mu _B}{t^2}) > 0$
$( - 4t + {t^2}(5{\mu _B} - \dfrac{1}{2})) > 0$
$\Rightarrow t( - 4 + t(5{\mu _B} - 0.5)) > 0$
$\Rightarrow ( - 4 + t(5{\mu _B} - 0.5)) > 0$
Since $t > 0$
$(5{\mu _B} - 0.5) > 0$
$\Rightarrow {\mu _B} > \dfrac{{0.5}}{5}$
$\therefore {\mu _B} > 0.1$

So the coefficient of friction between block B and ground must be greater than $0.1$.

Note: It should be remembered that friction is always not the force which stops the motion rather it may sometimes act as a cause of motion, example of a rotating sphere on an inclined plane here due to friction torque exerts on the sphere and hence it rolls down an inclined plane. Also friction is a blessing in disguise because it helps to have a control over block, if there were no friction then the body could not be stopped eventually and it would be harmful.