Question

What is the work to be done to increase the velocity of a car from $30km/hr$ to $60km/hr$ if the mass of the car is $1500kg$ ?

Answer 1

Hint:Work is defined as scalar or dot product of force applied and displacement moved due to that force. However, to solve this question we will use the work-energy principle in which work can be defined as change in kinetic energy.

Step by step solution:
Given from the question initial velocity ( ${v_1}$ ) = $30km/hr$
Final velocity ( ${v_2}$ ) = $60km/hr$
Mass of the car (m) = $1500kg$
Converting velocity in S.I unit in $m/s$ . We know that $1km\, = \,1000m$ and $1\,{\text{hour = 3600 seconds}}$ . So initial velocity ( ${v_1}$ ) = $\dfrac{{30\, \times \,1000}}{{3600}}\,m/s$
Solving above equation we get ( ${v_1}$ ) = $\dfrac{{25}}{3}\,m/s$
Similarly converting final velocity in $m/s$ . We can write ( ${v_2}$ ) = $\dfrac{{60\, \times \,1000}}{{3600}}\,m/s$
Solving above equation we get ( ${v_2}$ ) = $\dfrac{{50}}{3}\,m/s$
Now as we know that work is also defined as change in kinetic energy of a body and writing work-energy principle we can write
Work done (W) = final kinetic energy – initial kinetic energy
Work done (W) = $\dfrac{1}{2}m{v_2}^2\, - \,\dfrac{1}{2}m{v_1}^2$
Work done (W) = $\dfrac{m}{2}\left( {{v_2}^2\, - \,{v_1}^2} \right)$
Substituting values of m, ${v_1}$ and ${v_2}$ in above equation we get
Work done (W) = $\dfrac{{1500}}{2}\left( {{{\left( {\dfrac{{50}}{3}} \right)}^2}\, - \,{{\left( {\dfrac{{25}}{3}} \right)}^2}} \right)$J
Work done (W) = $\dfrac{{1500}}{2}\, \times \,\left( {\dfrac{{2500 - 625}}{9}} \right)$ J
Upon solving we can finally write Work done (W) = $156250$ Joule
So final answer on total work done (W) = $156.25$ kJ.

Note:We should also remember that S.I unit of work is Joule which is represented by J. For a force acting on a moving body amount of work is defined as
Work = force $ \times $ distance $ \times $ cosine of the angle between force and direction of motion and mathematically expressed as
Work = $Fd\cos \theta $
So, we can also infer that if $\theta $ is acute (less than 90 degree) work done is positive. For example, if we pull any object and it moves. Work done is zero in the case of an object moving in a circular orbit since $\theta $ is 90 degrees in that case. Work can also be negative if $\theta $ is obtuse, for example if a balloon is going upward against gravity.