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

Answer: (C)

5J

Answer 2

The work done against the frictional force can be calculated using the formula:

$\text{Work} = \mu_k \times N \times d$

Where:
- $\mu_k = 0.1$ (the coefficient of kinetic friction),
- $N = mg \cos \theta$ (the normal force),
- $d = 10 \, m$ (the distance moved along the inclined plane).

First, we find the normal force:

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

Now we can calculate the work done against friction:

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