A force F at an angle 𝜃𝜃 above the horizontal is used to pull a heavy suitcase of weight mg a distance d along a level floor at constant velocity. The coefficient of friction between the floor and the suitcase is 𝜇𝜇. The work done by the frictional force is:
Question
A force F at an angle 𝜃𝜃 above the horizontal is used to pull a heavy suitcase of weight mg a distance d along a level floor at constant velocity. The coefficient of friction between the floor and the suitcase is 𝜇𝜇. The work done by the frictional force is:
Solution
The work done by a force is given by the equation W = Fdcos(θ), where W is the work done, F is the force, d is the distance, and θ is the angle between the force and the direction of motion.
In this case, the frictional force is acting in the opposite direction to the motion of the suitcase. Therefore, the angle θ is 180 degrees. The cosine of 180 degrees is -1.
The frictional force can be calculated using the equation F_friction = μN, where μ is the coefficient of friction and N is the normal force. In this case, the normal force is equal to the weight of the suitcase, mg.
Therefore, the frictional force is F_friction = μmg.
Substituting this into the work done equation gives:
W = μmgdcos(180) = -μmgd
Therefore, the work done by the frictional force is -μmgd. The negative sign indicates that the work is done against the direction of motion.
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