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