Question

Calculate the work done by the frictional force in pulling a mass of \[50\;kg\] for a distance of \[100\;m\] on a road. The limiting coefficient of friction for a road is?

Answer 1

Hint: Whenever a body is moved along a surface, a frictional force equal to the product of normal force (force due to gravitation) and limiting coefficient of friction of that surface, acts on the body along a direction opposite to the direction of motion. Naturally, the coefficient of friction is different for surfaces of different material, based on the degree of their roughness.

Complete step by step solution: -
We have been given the mass of the body, and the distance for which it has been pulled.
Mass ($m$) $ = 50kg$
Displacement ($d$) $ = 100m$
Let us first calculate the normal force, that is, the force due to gravitation on this body.
Normal force is given by, $N = m \times g$, where $m$is the mass of the body and $g$ is the acceleration due to gravity, the value of which is usually taken as $10m/{s^2}$.
$\Rightarrow N = mg = 50 \times 10 = 500N$
Now, we have the frictional force given by, $F = \mu N$, where $\mu $denotes the limiting coefficient of friction and $N$is the normal force.
We have not been given the value of $\mu $here, so we will find the answer in terms of $\mu $.
$\Rightarrow F = \mu N = 500\mu $
Now, we know, work done is the product of force applied and the displacement along the direction of that force.
$W = Fd\cos \theta $, where $F$is the force applied, $d$is the displacement and $\theta $is the angle between the two.
Here, the body is moving opposite to the direction of the frictional force.
$\Rightarrow \theta = 180^\circ $
So, the work done would be:
$W = Fd\cos \theta = 500\mu \times 100 \times \cos (180^\circ ) = 500\mu \times 100 \times ( - 1) = - 50000\mu J = - 50\mu kJ$
So, the work done by the frictional force in the given displacement of the body is $ - 50\mu kJ$.

Note: -
The coefficient of friction is an important value, as it is different for different materials and surfaces. We have obtained a general value, assuming the value of limiting coefficient of friction to be $\mu $.
Another thing to be kept in mind in this type of problems, is that the displacement is against the force applied by friction, as friction applies a resistance to this displacement. The work done by the frictional force is thus always negative.