Question

A body and tackle with a velocity ratio of 5 is used to raise a mass of 50kg through a vertical distance of 20n at a steady rate if the effort is equal to 80n. Determine (a) distance moved by effort (b)the workdone by the effort in lifting the load (c)the losses in energy involve the operating in machine

Answer 1

a. Distance moved by effort = 1 meter

b. Work done by effort = 80 Joules

c. Losses in energy = 18 Joules

How to determine the parameters

a. Note that velocity ratio is defined as the ratio of a distance through which any part of a machine moves to the distance moved by its parts

Velocity ratio = distance moved by the effort/distance moved by the body

Where velocity ratio = 5

vertical distance = 20cm

Distance moved by effort = 5 × 0. 20 = 1 meter

b. Formula for work done is given as ;

Work done = Effort × distance moved by effort

Work done = 80 × 1

Work done = 80 Newtons

c. First, let's find the work output

Work output = m g h

Work output = 50 × 9.8 × 0. 20

Work output = 98 Joules

Losses in energy = Work output - work done

Losses in energy = 98 - 80

Losses in energy = 18 Joules

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