Question

A machine which is 75% efficient uses 12 Joules of energy in lifting up a 1kg mass through a certain distance. The mass is then allowed to fall through that distance. What will be its velocity at the end of its fall?
A) $\sqrt {32} m/s$
B) $\sqrt {24} m/s$
C) $\sqrt {18} m/s$
D) $\sqrt 9 m/s$

Answer 1

Hint: The work done on the mass will be stored as potential energy which can be used to find the height till which the mass is raised and then we can use the third equation of motion to find the velocity.
$
  W = mgh \\
  {v^2} - {u^2} = 2as \\
 $ where a would be equal to g (acceleration due to gravity) and s would be equal to h.

Complete step by step answer:We are given that the machine is 75% efficient. Therefore the work done by machine ,say W, will be equal to
$
  W = 75\% \times 12J \\
   \Rightarrow W = \dfrac{{75}}{{100}} \times 12 \\
   \Rightarrow W = 9J \\
 $
Now, we know that the work done would be stored in form of potential energy (energy of a body due to the virtue of its height). Let the mass of the object be m, let the acceleration due to gravity be g and let the height that the mass is raised to be h. Then,
Work done= Potential Energy
$
  W = P.E. \\
   \Rightarrow W = mgh \\
   \Rightarrow 9 = 1 \times 9.8 \times h \\
   \Rightarrow h = 0.918m \\
 $
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
$
  {v^2} - {u^2} = 2as \\
   \Rightarrow {v^2} - 0 = 2gh \\
   \Rightarrow {v^2} = 2 \times 9.8 \times 0.918 \\
   \Rightarrow v = \sqrt {18} m/s \\
 $
So, the correct answer is C) $\sqrt {18} m/s$ .

Note:We can use the energy conservation method to solve the question i.e. the potential energy gets converted to kinetic energy. Mathematically,
$
  mgh = \dfrac{1}{2}m{v^2} \\
   \Rightarrow 9 = \dfrac{1}{2} \times 1 \times {v^2} \\
   \Rightarrow v = \sqrt {18} m/s \\
 $