Question

A machine which is 75 percent efficient uses 12 J 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?
$A.\sqrt {18} m{s^{ - 1}} \\\ B.\sqrt {36} m{s^{ - 1}} \\\ C.\sqrt 8 m{s^{ - 1}} \\\ D.\sqrt 9 m{s^{ - 1}} \\\$

Answer 1

The law of conservation of energy states that the total energy of an isolated system remains constant. This implies that the total kinetic energy of a system is equal to the total potential energy of the system.
KE = PE

Complete step by step answer:
The work-energy theorem states that the total amount of work done on a system is stored as energy in the system.
In this case, when the machine lifts the mass through some distance in the air, there is work done. This work done, is stored as potential energy in the system.
When this mass is released from the height, the potential energy is converted to kinetic energy since the net total of the energy in the system remains constant.
Here, the machine uses 12J of energy.
Here, the efficiency given is, $\eta = 75\%$
Thus, actual work done, $W = 75\% \times 12J = 0.75 \times 12 = 9J$
Work done, $W = F \times s = mgh$
Substituting the values, we get the height at which the mass is lifted.
$W = mgh \\\ 9 = 1 \times 9.81 \times h \to \left( {g = 9.81m{s^{ - 2}}} \right) \\\ h = \dfrac{9}{{9.81}} = 0.917m \\\$
As per conservation of energy, we have –
$K.E = P.E \\\ \dfrac{1}{2}m{v^2} = mgh \\\$
Substituting,
$\dfrac{1}{2} \times 1 \times {v^2} = 1 \times 9.81 \times 0.917 \\\$
Solving,
${v^2} = 1 \times 9.81 \times 0.917 \times 2 \\\ {v^2} = 17.99 \sim 18 \\\ v = \sqrt {18} m{s^{ - 1}} \\\$
Hence, the correct option is Option A.

Note: In this question, the height through which the mass lifted is calculated from the potential energy by using equation: $PE = mgh$
However, the potential energy can be directly substituted in the equation $K.E = P.E$ instead of calculating the height separately.
$K.E = P.E = 9J$