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Question: A lifting machine, having an efficiency of 80% uses 2500 J of energy in lifting a 10 kg load over a ...

A lifting machine, having an efficiency of 80% uses 2500 J of energy in lifting a 10 kg load over a certain height. If the load is now allowed to fall through that height freely, its velocity at the end of the fall will be (Take acceleration due to gravity as 10m/s210m/{s^2}).
(A) 10ms110m{s^{ - 1}}
(B) 15ms115m{s^{ - 1}}
(C) 20ms120m{s^{ - 1}}
(D) 25ms125m{s^{ - 1}}

Explanation

Solution

In this problem,we are going to apply the concept of kinetic energy which is directly proportional to the mass of the object and to the square of its velocity and it can also be defined as the work needed to move a body of a given mass from rest to its stated velocity.

Formula used:
Kinetic energy=12m v2 = \dfrac{1}{2}m{\text{ }}{v^2}
Here,m = mass and v = velocity.

Complete step by step answer:
Given that potential energy given is 2500 J.
Energy is directly proportional to the mass of the object and to the square of its velocity. If the mass has units of kilograms and the velocity of meters per second, the kinetic energy has units of kilograms-meters squared per second squared.
Kinetic energy=12m v2 = \dfrac{1}{2}m{\text{ }}{v^2}
Energy stored in potential energy = 0.8×2500J0.8 \times 2500J
m = 10 kg
This energy is converted to Kinetic energy, that is,
Kinetic energy =12m v2 = \dfrac{1}{2}m{\text{ }}{v^2}
0.8×2500=12×10×v2\Rightarrow 0.8 \times 2500 = \dfrac{1}{2} \times 10 \times {v^2}
2000=5×v2 20005=v2 400=v2 400=v v=20ms1  \Rightarrow 2000 = 5 \times {v^2} \\\ \Rightarrow \dfrac{{2000}}{5} = {v^2} \\\ \Rightarrow 400 = {v^2} \\\ \Rightarrow \sqrt {400} = v \\\ \therefore v = 20m{s^{ - 1}} \\\
From the above calculations, we can conclude that the velocity at the end of the fall will be 20ms120 m{s^{ - 1}}, when a lifting machine having an efficiency of 80% uses 2500 J of energy in lifting a 10 kg load over a certain height.

Thus, the correct answer is option C.

Note: Here, one thing to be noted is that the potential energy and kinetic energy are two different kinds of energies. Kinetic energy is the energy required for an object or particle needed for the motion. On the application of the net force on an object, the object speeds up and consequently generates this energy. Kinetic energy is directly proportional to the mass of the object and to the square of its velocity which can be written as,
Kinetic energy =12m v2 = \dfrac{1}{2}m{\text{ }}{v^2}
Potential energy is energy an object has because of its position relative to some other object. The formula for potential energy depends on the force acting on the two objects. For the gravitational force the formula is
PotentialEnergy=mgh\Rightarrow Potential Energy = mgh
Where m = mass in kilograms,
G = acceleration due to gravity.