A block of mass 2·0 kg slides on a rough horizontal plane surface. Let

A block of mass 2·0 kg slides on a rough horizontal plane surface. Let the speed of the block at a particular instant is 10 m/s. It comes to rest after travelling a distance of 20 m. Which one of the following could be the magnitude of the frictional force ?

10 N
20 N
40 N
50 N
This question was previously asked in
UPSC NDA-2 – 2024
Given mass m = 2.0 kg, initial velocity u = 10 m/s, final velocity v = 0 m/s, distance s = 20 m. The block is brought to rest by the frictional force. We can calculate the work done by friction using the work-energy theorem, which states that the net work done on an object is equal to its change in kinetic energy. Work done by friction (W_friction) = Change in KE = (1/2)mv² – (1/2)mu². W_friction = (1/2)(2.0 kg)(0 m/s)² – (1/2)(2.0 kg)(10 m/s)² = 0 – 100 J = -100 J. Since the frictional force (F_friction) opposes the motion, the work done by friction is -F_friction * s. So, -F_friction * 20 m = -100 J. This gives F_friction = 100 J / 20 m = 5 N.

However, 5 N is not among the options. Let’s check the options against possible work done. If the force was 10N, the work done by friction over 20m would be -10N * 20m = -200 J. This would correspond to a change in KE of -200 J, meaning the initial KE would have to be 200 J. (1/2) * 2kg * u² = 200 J => u² = 200 => u = sqrt(200) ≈ 14.14 m/s.
If the distance was 10m instead of 20m, and the force was 10N, the work done would be -10N * 10m = -100J, matching the initial KE calculated from u=10m/s.
Given the discrepancy between the calculated value (5 N) and the provided options, and the common structure of such problems, it is highly probable there is a typo in the question’s distance value, and it was intended to be 10 m, leading to a force of 10 N (Option A). Alternatively, the options might be based on a different initial speed or mass. Assuming the question and options are from a single source and intended to have a correct answer among the choices, and that Option A is the intended answer, the underlying scenario would involve a force of 10N. While the provided data leads to 5N, Option A (10N) is the most plausible intended answer given the multiple choices.

The work-energy theorem states that the net work done on an object equals its change in kinetic energy. Frictional force does negative work as it opposes motion.
The magnitude of kinetic friction is often modeled as F_friction = μ_k * N, where μ_k is the coefficient of kinetic friction and N is the normal force. For a horizontal surface, N = mg.
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