An external force of 1 N is applied on the block of 1 kg as shown in the figure. The magnitude of the friction force F_s is (where, μ  = 0.3, g = 10 m/s²):

Concept:

To determine the magnitude of the friction force (\( F_s \)) when an external force of 1 N is applied to a 1 kg block, we need to use the formula for frictional force:

Given:

• \( \mu \) is the coefficient of friction
• \( N \) is the normal force

Values:

• \( \mu = 0.3 \)
• Mass of the block \( m = 1 \, \text{kg} \)
• Gravitational acceleration \( g = 10 \, \text{m/s}^2 \)
• Applied force \( F = 1 \, \text{N} \)

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First, we calculate the normal force \( N \):

\( N = m \cdot g = 1 \, \text{kg} \times 10 \, \text{m/s}^2 = 10 \, \text{N} \)

Next, we calculate the frictional force \( F_s \):

\( F_s = \mu \cdot N = 0.3 \times 10 \, \text{N} = 3 \, \text{N} \)

Since the frictional force is 3 N, and the applied force is 1 N, the block does not move because the applied force is less than the maximum static friction force.

Hence, the actual friction force will be equal to the applied force (since the block is not moving):

\( F_s = 1 \, \text{N} \)

Therefore, the magnitude of the friction force is:

4) 1 N