A block of mass 2 kg, moving with the initial speed of 3 m/s comes to

A block of mass 2 kg, moving with the initial speed of 3 m/s comes to rest on a rough horizontal surface after travelling a distance of 3 m. The magnitude of the frictional force is :

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9 N

3 N

18 N

1 N

Answer is Right!

Answer is Wrong!

This question was previously asked in

UPSC NDA-1 – 2024

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We can use the work-energy theorem, which states that the net work done on an object is equal to its change in kinetic energy. The block starts with an initial kinetic energy and comes to rest, meaning its final kinetic energy is zero. The change in kinetic energy is the final KE minus the initial KE. Initial KE = (1/2) * m * v_i^2 = (1/2) * 2 kg * (3 m/s)^2 = 9 J. Final KE = (1/2) * m * v_f^2 = (1/2) * 2 kg * (0 m/s)^2 = 0 J. Change in KE = 0 – 9 = -9 J. The only force doing work in the horizontal direction is the frictional force (f), opposing the motion. The work done by friction is W_friction = f * d * cos(180°) = -f * d, where d is the distance (3 m). According to the work-energy theorem, W_friction = Change in KE. So, -f * 3 m = -9 J. Solving for f, we get f = (-9 J) / (-3 m) = 3 N. The magnitude of the frictional force is 3 N.

– Work-Energy Theorem: Net Work Done = Change in Kinetic Energy.
– Initial Kinetic Energy = (1/2)mv_i^2.
– Final Kinetic Energy = (1/2)mv_f^2.
– Work done by friction = – f * d.

Alternatively, one could calculate the deceleration using kinematics (v_f^2 = v_i^2 + 2ad) and then find the force using Newton’s second law (F=ma). 0^2 = 3^2 + 2 * a * 3 => 0 = 9 + 6a => 6a = -9 => a = -1.5 m/s^2. The magnitude of acceleration is 1.5 m/s^2. The net force in the direction of motion is the frictional force (acting opposite to initial motion). F_net = m * a = 2 kg * (-1.5 m/s^2) = -3 N. The magnitude of the force is 3 N.

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