Question

An iron block of mass $5\,{\text{kg}}$ is kept in a trolley. If the trolley is being pushed with an acceleration of $5\,{\text{m/}}{{\text{s}}^2}$, what will be the force of friction between the block and the trolley surface. (Take the coefficient of static friction between the block and the surface to be 0.8).
A. zero
B. $5\,{\text{N}}$
C. $4\,{\text{N}}$
D. $25\,{\text{N}}$

Answer 1

Use the formula for the static frictional force between an object and a surface. Also use the expression for Newton’s second law of motion. First determine the maximum value of the static frictional force between the block and trolley. Also determine the force on the block due to motion of the trolley. From the values of these two forces, determine the value of the frictional force between block and trolley for the block to be stationary.

Formulae used:
The force of static friction ${F_S}$ between an object and a surface is given by
${F_S} = {\mu _S}mg$ …… (1)
Here, ${\mu _S}$ is the coefficient of static friction between the object and surface, $m$ is mass of the object and $g$ is acceleration due to gravity.
The expression for Newton’s second law of motion is
${F_{net}} = ma$ …… (2)
Here, ${F_{net}}$ is net force acting on the object, $m$ is mass of the object and $a$ is acceleration of the object.

Complete Step by Step Answer:
We have given that an iron block of mass $5\,{\text{kg}}$ is kept in a trolley and the acceleration of the trolley is $5\,{\text{m/}}{{\text{s}}^2}$.
$m = 5\,{\text{kg}}$
$a = 5\,{\text{m/}}{{\text{s}}^2}$
The coefficient of static friction between the block and the trolley is 0.8.
${\mu _S} = 0.8$
We have asked to calculate the force of friction between the block and the trolley. Let us first determine the maximum value of the static friction force between the block and the trolley.
Substitute $0.8$ for ${\mu _S}$, $5\,{\text{kg}}$ for $m$ and $10\,{\text{m/}}{{\text{s}}^2}$ for $g$ in equation (1).
${F_S} = \left( {0.8} \right)\left( {5\,{\text{kg}}} \right)\left( {10\,{\text{m/}}{{\text{s}}^2}} \right)$
$\Rightarrow {F_S} = 40\,{\text{N}}$
Hence, the maximum value of force of static friction between the block and the trolley is $40\,{\text{N}}$.

Let us now calculate the force on the block due to acceleration of the trolley.For this, consider that the frictional force between the block and trolley is zero. Since the block moves in the opposite direction due to acceleration of the trolley. The acceleration of the block is $- 5\,{\text{m/}}{{\text{s}}^2}$.Substitute $5\,{\text{kg}}$ for $m$ and $- 5\,{\text{m/}}{{\text{s}}^2}$ for $a$ in equation (2).
$F = \left( {5\,{\text{kg}}} \right)\left( { - 5\,{\text{m/}}{{\text{s}}^2}} \right)$
$\therefore F = - 25\,{\text{N}}$
Hence, the force on the block due to motion of the trolley is $25\,{\text{N}}$.The negative sign indicates that the motion of the block and acceleration of the trolley has opposite directions.Since the block is stationary in the trolley, the static frictional force must be equal to the force on the block due to motion of the trolley.Therefore, the frictional force between the block and trolley is $25\,{\text{N}}$.

Hence, the correct option is D.

Note: The students may calculate the maximum static frictional force between the iron block and trolley and get confused that the correct answer is not given in the options. But we need to calculate the minimum frictional force for the iron block to be stationary on the trolley. So this force must be equal to the force on block due to motion of the trolley in the opposite direction to keep the block at rest.