Question

A block of mass 1 kg slides down a curved track that is one quadrant of a circle of radius 1 m. Speed of the block at the bottom is 2 m/s. Work done by the frictional force on the block when it reaches at the bottom is

• A. 8 J
• B. -8 J
• C. 4 J
• D. 0 J

Hint:

Work done by non-conservative forces = change in mechanical energy.

Answer 1

The Correct Option is B

To solve this problem, we use the principle of conservation of energy. The block slides down a frictional track, which does work against friction. The initial potential energy is converted into kinetic energy and work done against friction.

Calculate the initial potential energy (PE) at the top of the track. The height is equal to the radius of the circle since it slides down one quadrant. Thus, \(h = 1 \, \text{m}\), the mass \(m = 1 \, \text{kg}\), and gravitational acceleration \(g = 9.8 \, \text{m/s}^2\).

\(\text{PE} = mgh = 1 \times 9.8 \times 1 = 9.8 \, \text{J}\)

Calculate the kinetic energy (KE) at the bottom. The speed at the bottom is given as \(v = 2 \, \text{m/s}\).

\(\text{KE} = \frac{1}{2}mv^2 = \frac{1}{2} \times 1 \times (2)^2 = 2 \, \text{J}\)

According to the work-energy principle, the work done by all non-conservative forces (here, friction) is equal to the change in mechanical energy of the system.

\(\text{Work done by friction} = \text{KE at bottom} - \text{Initial PE}\)

\(= 2 - 9.8 = -7.8 \, \text{J}\)

The correct answer is option:

-8 J 
 

. Note: Since the question likely rounds to a more precise integer, the actual work done is close to -8 J.