Problem#1
A block of mass 2.50 kg is pushed 2.20 m along a frictionless horizontal table by a constant 16.0-N force directed 25.0° below the horizontal. Determine the work done on the block by (a) the applied force, (b) the normal force exerted by the table, and (c) the gravitational force. (d) Determine the total work done on the block.
Answer:
Given: m = 2.50 kg, Δr = 2.20 m, F = 16.0 N and θ = 25.00, then
(a) the work done on the block by the applied force,
W = F.Δr cosθ
W = (16.0 N)(2.20 m) cos25.00 = 31.9 J
(b), (c) The normal force and the weight are both at 90° to the displacement in any time interval.
Both do 0 work.
(d) the total work done on the block is
Wtotal = 31.9 J + 0 + 0 = 31.9 J
Problem#2
A shopper in a supermarket pushes a cart with a force of 35.0 N directed at an angle of 25.0° downward from the horizontal. Find the work done by the shopper on the cart as he moves down an aisle 50.0 m long.
Answer:
Given: Δr = 50.0 m, F = 35.0 N and θ = 25.00, then
the work done by the shopper on the cart as he moves down an aisle 50.0 m long, given by
W = F.Δr cosθ
W = (35.0 N)(50.0 m) cos25.00 = 1.59 kJ
Problem#3
Batman, whose mass is 80.0 kg, is dangling on the free end of a 12.0-m rope, the other end of which is fixed to a tree limb above. He is able to get the rope in motion as only Batman knows how, eventually getting it to swing enough that he can reach a ledge when the rope makes a 60.0° angle with the vertical. How much work was done by the gravitational force on Batman in this maneuver?
Answer:
The force of gravity on Batman is
mg = (80 kg)(9.80 m/s2) = 784 N down.
Only his vertical displacement contributes to the work gravity does. His original y-coordinate below the tree limb is ri = –12.0 m.
His final y-coordinate is
rf = –12 m cos60.00 = –6.00 m
The work done by gravity is
W = F.Δr cosθ
W = (35.0 N)(–6.00 m + 12.0 m) cos1800 = –4.70 kJ
Problem#4
A raindrop of mass 3.35 x 10-5 kg falls vertically at constant speed under the influence of gravity and air resistance. Model the drop as a particle. As it falls 100 m, what is the work done on the raindrop (a) by the gravitational force and (b) by air resistance?
Answer:
(a) the work done on the raindrop by the gravitational force is
W = F.Δr cosθ = mgΔr cos00
W = (3.35 x 10-5 kg)(100 m) cos00 = 0.0328 J
(a) the work done on the raindrop by by air resistance is
W = F.Δr cosθ = fΔr cos1800
W = (3.35 x 10-5 kg)(100 m) cos1800 = –0.0328 J
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