A 100kg box is pushed along a road with force of 500N. If the box moves with a uniform velocity, the coefficient of friction between the box and the road is

Assessment: JAMB UTME - Physics - 2004 Subject: Physics

Question 1 Report

A 100kg box is pushed along a road with force of 500N. If the box moves with a uniform velocity, the coefficient of friction between the box and the road is
Answer Details
To find the coefficient of friction between the box and the road, we need to use the formula for frictional force, which is given by: f_friction = coefficient of friction x normal force. Since the box is moving with a uniform velocity, we know that the net force acting on the box is zero. The force pushing the box forward is balanced by the force of friction acting in the opposite direction. Therefore, we can set up an equation to find the force of friction acting on the box: f_push - f_friction = 0 where f_push is the force pushing the box forward, and f_friction is the force of friction acting in the opposite direction. Substituting the values given in the problem, we get: 500N - f_friction = 0 Solving for f_friction, we get: f_friction = 500N Now, we can use the formula for frictional force to find the coefficient of friction: f_friction = coefficient of friction x normal force The normal force is the force exerted by the road on the box, which is equal to the weight of the box (since the box is not accelerating): normal force = weight of box = mg = 100kg x 9.8 m/s^2 = 980N Substituting this value and the value we found for the force of friction into the formula for frictional force, we get: 500N = coefficient of friction x 980N Solving for the coefficient of friction, we get: coefficient of friction = 500N / 980N = 0.51 Therefore, the coefficient of friction between the box and the road is approximately 0.5, which is the fourth option.

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