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?

Answer 1

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
$$\begin{gathered} W=75\mathrm{\%}\times 12J \\ \Rightarrow W=\frac{75}{100}\times 12 \\ \Rightarrow W=9J \end{gathered}$$

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
$$\begin{gathered} W=P.E. \\ \Rightarrow W=mgh \end{gathered}$$

Where,

$m\to$ the mass of the object

$g\to$ the acceleration due to gravity

$h\to$ the height that the mass is raised to be

$$\begin{gathered} \Rightarrow 9=1\times 9.8\times h \\ \Rightarrow h=0.918m \end{gathered}$$

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
$$\begin{gathered} v^2-u^2=2as \\ \Rightarrow v^2-0=2gh \\ \Rightarrow v^2=2\times 9.8\times 0.918 \\ \Rightarrow v=\sqrt{18}m/s \end{gathered}$$

Thus, the velocity at the end of its fall is $\sqrt{18}m/s$.