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

The Correct Option is B

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 \( 147 \, \text{N} \).

Answer 2

Step 1: Given data.
Mass of the box, m = 50 kg
Coefficient of kinetic friction, μ_k = 0.3
Acceleration due to gravity, g = 9.8 m/s²

Step 2: Formula for force of kinetic friction.
The kinetic frictional force is given by:
\[ F = \mu_k N \]
For a horizontal surface, the normal force (N) is equal to the weight of the object:
\[ N = m g \]

Step 3: Substitute the values.
\[ F = 0.3 \times 50 \times 9.8 \] \[ F = 0.3 \times 490 = 147 \, \text{N} \]

Step 4: Final Answer.
The force of kinetic friction is:
\[ \boxed{147 \, \text{N}} \]

Final Answer: 147 N