Question

A 250 g block slides on a rough horizontal table. Find the work done by the frictional force in bringing the block to rest if it is initially moving at a speed of . If the friction coefficient between the table and the block is 0.1, how far does the block move before coming to rest?

Answer 1

In this solution, we will use the work-energy theorem which tells us that the total work done will be equal to the change in kinetic energy of the block. The work done by the frictional force will be the product of the frictional force and the distance moved by the object.

Formula used: In this solution, we will use the following formulae:
- Frictional force: ${F_f} = \mu mg$ where $\mu$ is the coefficient of friction, $m$ is the mass of the block, $g$ is the gravitational acceleration.
- Kinetic energy of the block: $K = \dfrac{1}{2}m{v^2}$ where $v$ is the velocity of the object.

Complete step by step answer:
In the situation given to us, a 250 g block slides on a rough horizontal table. The work done by the friction force in making the block come to rest will be equal to the change in kinetic energy of the block.
So, the work done by the friction force will be the product of the frictional force and the distance travelled by the block as
${F_f}.d = \Delta K$
The friction force acting on the object is calculated as
${F_f} = \mu mg$
The change in kinetic energy of the block will be
$\Delta K = \dfrac{1}{2}m({v^2} - {u^2})$
Substituting these values in the work-energy theorem equation, we get
$\mu mg.d = \dfrac{1}{2}m({v^2} - {u^2})$
Since the final velocity of the block will be zero as it comes to rest, the initial velocity is $u = 40\;cm/s$ , the coefficient of friction is $\mu = 0.1$ , we can cancel out the mass on both sides of the equation and write
$0.1 \times 10 \times d = \dfrac{1}{2} \times 0.16$
$\Rightarrow d = 0.08\,m$
Or
$d = 8\,cm$
Hence the block will travel for 8 cm before coming to rest.

Note:
We can directly calculate the work done by the friction force as the friction force acting on the block will be constant irrespective of its velocity. When using the work-energy balance theorem equation, we must ensure that all the variables have units in the same system.