Answer:
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
