Question

Certain force acting on a $20kg$ mass changes its velocity from $5m{s^{ - 1}}$ to $2m{s^{ - 1}}$. Calculate the work done by the force. If time taken for this change is $2\sec$. Calculate its power.

Answer 1

This problem is taken from the topic introduction of work and energy.Firstly, We have to calculate the initial and final kinetic energy and then find the subtraction of the two to find the work done. In the second part we are asked for the power which is multiplication of the work done and time.

Complete step by step answer:
We are given that, mass of the body $\left( m \right)$= $20kg$.
Initial velocity of the body $\left( u \right)$= $5m{s^{ - 1}}$.
Final velocity of the body $\left( v \right)$= $2m{s^{ - 1}}$.
Using the work formula,
Work done = change in the kinetic energy
Finding, the initial kinetic energy$\left( {{E_i}} \right)$ $= \dfrac{1}{2}m{u^2}$
${E_i} = \dfrac{1}{2} \times 20 \times {\left( 5 \right)^2}$
$\Rightarrow {E_i} = 250J$
Now, the final kinetic energy $\left( {{E_f}} \right)$ $= \dfrac{1}{2}m{v^2}$
${E_f} = \dfrac{1}{2} \times 20 \times {\left( 2 \right)^2}$
$\Rightarrow {E_f} = 40J$
Now, the work done= final kinetic energy- initial kinetic energy
W = {E_f} - {E_i} \\\ \Rightarrow W= 250 - 40 \\\ \Rightarrow W = 210J \\\
Hence, work done by the object is $210J$.
Now, we have to find the power $\left( P \right)$ for change of time $\left( t \right)$,
$P = \dfrac{W}{t}$
$\therefore P = \dfrac{{210}}{2} = 105Watt$

Hence, power is $105Watt$.

Note: Units should be written properly. We have solved the problem from the work-energy theorem which states that total work done is given by the change in kinetic energy. And the second statement is that work done is given by the addition of all the forces acting on the particle.