Question

A machine which is 72 percent efficient, uses 36 joules of energy in lifting up 1kg mass through a certain distance. The mass is then allowed to fall through that distance. The velocity at the end of its fall is
A. $6.6\,m{s^{ - 1}}$
B. $7.2\,m{s^{ - 1}}$
C. $8.1\,m{s^{ - 1}}$
D. $9.2\,m{s^{ - 1}}$

Answer 1

According to conservation of energy, energy can neither be created nor be destroyed but can be transferred from one form to another. Using conservation of energy in the initial and final state will lead to the answer.

Complete step by step answer:
The machine lifts the object to a specified height, let that height be h, and then leaves the object to fall freely from that height (h). As the object is raised up by the machine, it gains the potential energy and free fall will generate the kinetic energy in the object.
Now according to conservation of energy:
Kinetic energy = Potential energy
The work done by machine in lifting the object is stored in substance as its potential energy, hence we can say,
Work done = Potential energy
The efficiency of machine is 72% i.e., actual work done by machine will be then,
$W = 72\% \,of\,36 = 72\% \times 36$
Where, W = Actual work done by machine = Potential energy stored in object
$W = \dfrac{{72}}{{100}} \times 36$

Evaluating above equation will give
$W = 25.92\,J$
$W$ = Potential energy (P.E.) = $mgh = 25.92\,J$
And according to energy conservation
Potential energy (P.E.) = Kinetic energy (K.E.)
$K.E. = \dfrac{1}{2}m{v^2}$
Equating potential energy and kinetic energy equal we will get,
$\dfrac{1}{2}m{v^2} = mgh = 25.92$

Putting value of mass in formula of K.E. in above equation will give,
$\dfrac{1}{2} \times 1 \times {v^2} = 25.92$
Evaluating above equation will give,
${v^2} = 25.92 \times 2 \\\ \Rightarrow {v^2} = 51.84$
$\Rightarrow v = \sqrt {51.84} \\\ \therefore v = 7.2\,m{s^{ - 1}}$
The velocity attained by an object after free falling the lifted height is $7.2\,m{s^{ - 1}}$.

Therefore, the correct answer is option B.

Note: Efficiency of the machine is the amount of work done by machine over the total input work on machine. Output work or work done by machine is always less than the input work. In general efficiency of machines is omitted but if mentioned it is necessary to apply it as neglecting will generate a quite different answer.