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 :
- 14.7 N
- 147 N
- 1.47 N
- 1470 N
The Correct Option is B
Approach Solution - 1
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 \( 147 \, \text{N} \).
Approach Solution -2
Mass of the box, m = 50 kg
Coefficient of kinetic friction, μk = 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
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