Please help me with b, c, and d. There are no angles.

a. 150 J

b. 150 J

c. 0 J

d. 0 J

Explanation:

The given parameters are;

The horizontal force with which the man pulls the canister, F = 50 N

The distance he moves the vacuum cleaner, d = 3.0 m

a. Work done, W = Force applied, F × Distance moved by the force, d

Therefore, for the work done by the 50 N force on the canister, we have;

W = 50 N × 3.0 m = 150 N·m = 150 J

b. Given that he pulls the canister at a constant speed, we have;

The acceleration of the canister, a = 0 m/s²

Therefore, the net force on the canister, = F -  = m × a

Where;

m = The mass of the canister

a = The acceleration of the canister

F = The applied force = 50 N

= The force of friction

∴ = m × a = m × 0 m/s² = 0 N

Therefore;

=  F - = 0 N

From which we have;

F = = 50 N (The applied force, F is equal to the force of friction,

The work done by friction = The force of friction × The distance in which the force of friction acts

∴ The work done by friction = × d - 50 N × 3.0 m = 150 J

The work done by friction = 150 J

c. The normal force, N acts perpendicular to the force of friction

The distance the canister moves in the perpendicular direction, = 0 m

∴ The work done by the normal direction = N × = N × 0 m = 0 J

The work done by the normal direction = 0 J

d. The vacuum weight, W, acts on the same line as the normal force but in the opposite direction to the normal force, N

Therefore, the weight, W, acts perpendicular to the line of motion of canister

The distance the canister moves in the direction of the weight, = 0 m

Therefore, the work done by the weight = W × = W × 0 m = 0 J

The work done by the weight = 0 J

force required to detach it is given by

charges are taken to be fundamental charge

now we have

so above is the minimum distance at which it will detach