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Question

A machine that is 75% efficient uses 12J of energy in lifting up a 1kg mass through a certain distance. The mass is then allowed to fall through that distance. What will its velocity be at the end of its fall?


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Solution

Step 1: Given data

The potential energy of a body = 75% of 12J

Mass = 1kg

Step 2: Finding work done by the machine

We are given that the machine is 75% efficient.

Therefore the work done by a machine, say W, will be equal to
W=75%×12JW=75100×12W=9J

Step 3: Finding the height at fall

Now, we know that the work done would be stored in the form of potential energy (energy of a body due to the virtue of its height).

Then, Work done= Potential energy
W=P.E.W=mgh

Where,

m the mass of the object

g the acceleration due to gravity

h the height that the mass is raised to be

9=1×9.8×hh=0.918m

Step 4: Finding the velocity at the end of the fall

Now, we assume the final velocity of the mass as v.

The initial velocity is zero as the mass is dropped, the acceleration would be the acceleration due to gravity and the distance would be the height that the mass is raised to.
Applying the third equation of motion on the mass, we get
v2u2=2asv20=2ghv2=2×9.8×0.918v=18m/s

Thus, the velocity at the end of its fall is 18m/s.


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