Question

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

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

Hint:

The force of kinetic friction is given by \( F_k = \mu_k \cdot N \), where \( N \) is the normal force and \( \mu_k \) is the coefficient of kinetic friction.

Answer 1

The Correct Option is B

Step 1: Formula for Kinetic Friction Force
The force of kinetic friction \( F_k \) is given by the formula: \[ F_k = \mu_k \cdot N \] where \( \mu_k \) is the coefficient of kinetic friction and \( N \) is the normal force.
Step 2: Calculate the Normal Force
For a horizontal surface, the normal force \( N \) is equal to the weight of the box, which is the mass \( m \) multiplied by the acceleration due to gravity \( g \): \[ N = m \cdot g = 50 \times 9.8 = 490 \, \text{N} \]
Step 3: Calculate the Force of Kinetic Friction
Now, substitute the values into the kinetic friction formula: \[ F_k = 0.3 \times 490 = 147 \, \text{N} \] Final Answer: The force of kinetic friction is 147 N.