A block of mass $5\, kg$ is placed at rest on a table of rough surface Now, if a force of $30\, N$ is applied in the direction parallel to surface of the table, the block slides through a distance of $50\, m$ in an interval of time $10 \,s$ Coefficient of kinetic friction is (given, $g =10 \,ms ^{-2}$ ):

Question

A cubic block of mass $ m $ is sliding down on an inclined plane at $ 60^\circ $ with an acceleration of $ \frac{g}{2} $, the value of coefficient of kinetic friction is:

Answer 1

To determine the coefficient of kinetic friction, we need to balance the forces acting on the block in the horizontal direction. Given data includes the mass of the block $ m = 5 \, \text{kg} $, applied force $ F = 30 \, \text{N} $, distance moved $ s = 50 \, \text{m} $, time interval $ t = 10 \, \text{s} $, and gravitational acceleration $ g = 10 \, \text{m/s}^2 $.

First, calculate the acceleration of the block using the equation of motion:

s = ut + \frac{1}{2} a t^2

Here, initial velocity $ u = 0 $ (since the block starts from rest), so the equation becomes:

50 = 0 \times 10 + \frac{1}{2} a (10)^2

Simplifying, we find:

50 = 50a

Thus, the acceleration a = 1 \, \text{m/s}^2.

Now, consider the forces. The net force acting in the horizontal direction can be calculated using Newton's second law:

F_{\text{net}} = m \times a = 5 \times 1 = 5 \, \text{N}

The frictional force $ F_{\text{friction}} $ opposes the applied force:

F_{\text{friction}} = F - F_{\text{net}} = 30 - 5 = 25 \, \text{N}

The frictional force is also given by:

F_{\text{friction}} = \mu_k \times N

Here, $ \mu_k $ is the coefficient of kinetic friction, and $ N $ is the normal force. Since the block slides horizontally on a surface:

N = m \times g = 5 \times 10 = 50 \, \text{N}

Substituting in the known values:

25 = \mu_k \times 50

Solving for $ \mu_k $, we find:

\mu_k = \frac{25}{50} = 0.50

Therefore, the coefficient of kinetic friction is 0.50.

Answer 2