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 :

Updated On: Mar 22, 2025
  • 14.7 N
  • 147 N
  • 1.47 N
  • 1470 N
Hide Solution
collegedunia
Verified By Collegedunia

The Correct Option is B

Solution and Explanation

Given:
- Mass of the box: m=50kg m = 50 \, \text{kg}
- Coefficient of kinetic friction: μk=0.3 \mu_k = 0.3
- Acceleration due to gravity: g=9.8m/s2 g = 9.8 \, \text{m/s}^2

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

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

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

Fk=μkN. F_k = \mu_k N.

Substituting the values:

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

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

Was this answer helpful?
0
0