(a) Explain the term work
(b) Draw a diagram of a pulley system with a velocity ratio of 5.
(c) A man pulls up a box of mass 70kg using an inclined plane of effective length 5m onto a platform 2.5m high at uniform speed. If the frictional force between the box and the plane is 100N, draw the diagram of all the forces acting on the box when in motion and calculate the
(i) minimum effort applied in pulling up the box
(ii) velocity ratio of the plane
(iii) mechanical advantage of the plane
(iv) efficiency of the plane
(v) energy lost in the system
(vi) work output of the man
(vii) total power developed by the man given that the time taken to raise the box onto the platform is 50s. (g = 10ms\(^{-2}\)).
(a) Work is done whenever a force moves a body through a certain distance in the direction of the force. It is measured in joule and it is a scalar quantity.
(i) Minimum effort (E) applied in pulling up the box = F = mgsin\(\theta\) = 100 + 700 sin30° = 100 + 700 x 0.5 = 100 + 350=450N.
(ii) Velocity ratio (V.R) of the plane = 1/sin\(\theta\) - 5/2.5 = 2
(iii) Mechanical advantage of the plane (M.A) = L/E = 1.56 .
(iv) Efficiency of the plane = \(\frac{M.A}{V.R}\) x 100%
= \(\frac{1.56}{2}\) x 100% = 78%
(v) Energy lost in the system = work input - work output = (450 x 5) - (700/2 x 5) = 2250 - 1750 = 550J.
(vi) Work output of the man = mgh = 70 x 10 x 2.5 = 1750J
(vii) Total power developed by the man
= \(\frac{\text{work done by effort}}{time}\) = \(\frac{450 \times 5}{50}\) = 45 watts
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