What is the work done on the basket by friction, tension force, and normal force?

Question

A basket of letters, having a mass of 5.00 kg was pulled to the right with a displacement of 2.50 m on the ground surface. A thick string is tied to the bottom of the basket. (a) What is the work done on the basket by friction, tension force, and normal force, and (b) what is the total work done on the basket? The coefficient of kinetic friction is 0.30. (16 pts)

Answer 1

W = Fd where "W" is work, "F" is the force in the direction of travel, and "d" is the distance of the travel.

The frictional force is calculated as F_F = μN where F_F is the frictional force, μ is the coefficient of friction, and "N" is the normal force.

Summing the forces in the y-direction: W - N = 0 (where W = weight = mg) therefore W = mg = (5.00)(9.81) = 49.05 N and since W = N, the N = 49.05 N

Using that to calculate the frictional force: F_F = μN = (0.30)(49.05) = 14.715 N

So, since the friction is in the direction opposite to the travel, W = -(14.715)(2.50) = -36.8 Nm or -36.8 j

You don't say if the basket is being pulled at constant velocity or if there is an acceleration so there is no way to know what the tension force is. Perhaps that fact is given in the problem somewhere. If you know the tension force, then the work done is (T)(2.5)

The total work done is the sum of those two.

The normal force does NO work because the motion is not in the direction of N.