Question

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 :

• A. 14.7 N
• B. 147 N
• C. 1.47 N
• D. 1470 N

Answer 1

Answer: (B)

147 N

Answer 2

Given:
- Mass of the box: $m = 50 \, \text{kg}$
- Coefficient of kinetic friction: $\mu_k = 0.3$
- Acceleration due to gravity: $g = 9.8 \, \text{m/s}^2$

Step 1: Calculate the Normal Force
The normal force $N$ acting on the box is equal to the weight of the box, given by:

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

Step 2: Calculate the Force of Kinetic Friction
The force of kinetic friction $F_k$ is given by:

$F_k = \mu_k N.$

Substituting the values:

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

Therefore, the force of kinetic friction is $147 \, \text{N}$.