A heavy box of mass 50 kg is moving on a horizontal surface. If co-efficient of kinetic friction between the box and horizontal surface is 0.3 then force of kinetic friction is :

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

Provided Data:
- Box mass: $ m = 50 \, \text{kg} $
- Kinetic friction coefficient: $ \mu_k = 0.3 $
- Gravitational acceleration: $ g = 9.8 \, \text{m/s}^2 $

Calculation of Normal Force
The normal force $ N $ equals the box's weight:

\[ N = mg = 50 \times 9.8 = 490 \, \text{N}. \]

Calculation of Kinetic Friction Force
The kinetic friction force $ F_k $ is calculated as:

\[ F_k = \mu_k N. \]

Substituting values:

\[ F_k = 0.3 \times 490 = 147 \, \text{N}. \]

Result: The kinetic friction force is $ 147 \, \text{N} $.

A heavy box of mass 50 kg is moving on a horizontal surface. If co-efficient of kinetic friction between the box and horizontal surface is 0.3 then force of kinetic friction is :

The Correct Option is B

Solution and Explanation

Top Questions on Friction

Questions Asked in JEE Main exam