Answer:
(a) 6000 J
(b) 6800 J
(c) 3.53
(d) 88.23%
Explanation:
(a) Work output: This is the work done by the pulley.
W₁ = L × x.................... Equation 1
Where W₁ = work output, L = weight of the load or force required by the pulley to lift the load, x = distance moved by the load.
Given: L= 1200 N, x = 5.00 m
Substituting these values into equation 1,
W₁ = 1200×5
W₁ = 6000 J
Thus the work output = 600 J
(b) Work input: This is the work done by the person pulling the rope in the pulley.
W₂ = E× y.................................. Equation 2
Where W₂ = work input, E = force applied to the pulley or Effort, y = length of the rope.
Given: F₂ = 340 N, d₂ = 20 m
Substituting these values into equation 2
W₂ = 340×20
W₂ = 6800 Joules
Thus the work input = 6800 J
(c) Mechanical advantage: This is the ratio of load to effort of a machine.
M.A = L/E.................... Equation 3
Where M.A = mechanical advantage, L = load, E = effort
Given: L = 1200 N, E = 340 N
Substituting into equation 3
M.A = 1200/340
M.A = 3.53.
Thus the mechanical advantage of the machine = 3.53
(d) Efficiency of a machine: This is the ratio of work output to work input expressed in percentage.
E(%) = (W₁/W₂)×100
E(%) = (6000/6800)×100
E(%) = 88.23%
Thus the percentage efficiency of the machine = 88.23%
