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What is the work to be done to increase the velocity of a car from 30km/h to 60km/h if the mass of the car is 1500kg?
Given,
Mass of the car, m = 1500 kg
Initial velocity of the car, u = 30 km/h = $\frac{30\times 1000m}{3600s}=\frac{25}{3}m/s$ [converted km/h to m/s]
Final velocity of the car, v = 60 km/h = $\frac{60\times 1000m}{3600s}=\frac{50}{3}m/s$ [converted km/h to m/s]
To find = Work done (W)
Solution:
According to the Work-Energy theorem or the relation between Kinetic energy and Work done - the work done on an object is the change in its kinetic energy.
So, Work done on the car = Change in the kinetic energy (K.E) of the car
= $Final\ K.E-Initial\ K.E$
$Work\ done, \ W =\frac{1}{2}m{v}^{2}-\frac{1}{2}m{u}^{2}$ $[\because K.E=\frac{1}{m}{v}^{2}, \ where, \ mass\ of\ the\ body=m,\ and\ the\ velocity\ with\ which\ the\ body\ is\ travelling=v]$
$W=\frac{1}{2}m[{v}^{2}-{u}^{2}]$ $[taking\ out\ common]$
Now, substituting the values-
$W=\frac{1}{2}\times 1500[(\frac{50}{3}{)}^{2}-(\frac{25}{3}{)}^{2}]$
$W=\frac{1}{2}\times 1500[(\frac{50}{3}+\frac{25}{3})(\frac{50}{3}-\frac{25}{3})]$ $[\because ({a}^{2}-{b}^{2})=(a+b)(a-b)]$
$W=\frac{1}{2}\times 1500\times \frac{75}{3}\times \frac{25}{3}$
$W=156250J$
Hence, the work to be done to increase the velocity of a car from 30km/h to 60km/h is 156250 joule, if the mass of the car is 1500 kg.
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