Question

A block of mass 1 kg is pushed up a surface inclined to horizontal at an angle of 60° by a force of 10 N parallel to the inclined surface as shown in the figure. When the block is pushed up by 10 m along the inclined surface, the work done against frictional force is:

• A. \( \sqrt{5} \, J \)
• B. \( 5 \times 10^3 \, J \)
• C. 5J
• D. \( 10 \, J \)

Answer 1

The Correct Option is C

The work performed against the frictional force is determined by the formula:

\[ \text{Work} = \mu_k \times N \times d \]

Definitions: - \( \mu_k = 0.1 \) (coefficient of kinetic friction), - \( N = mg \cos \theta \) (normal force), - \( d = 10 \, m \) (distance traversed on the inclined plane).

The normal force is calculated first:

\[ N = mg \cos(60^\circ) = 1 \times 10 \times \frac{1}{2} = 5 \, N. \]

Subsequently, the work done against friction is computed:

\[ \text{Work} = \mu_k \times N \times d = 0.1 \times 5 \times 10 = 5 \, J. \]