What is the work to be done to increase the velocity of a car from 30km/h to 40km/h if the mass of the car is 1500 kg?

Text Solution

Answer : A::B

Solution : Here, `u = 30km//h = (30xx(5)/(18))m//s = (25)/(3)m//s (1km//h = (5)/(18) m//s)` <br> `upsilon = 60km//h = (60xx(5)/(18))m//s= (50)/(3)m//s` <br> m = 1500kg <br> According to work-energy theorem, <br> `W = (1)/(2)m upsilon^2 - (1)/(2)mu^2 = (1)/(2)m(upsilon^2 - u^2)` <br> or `W =(1)/(2)xx1500kg[((50)/(3)m//s)^2 - ((25)/(3)m//s)^2]` <br> `=750kg[((2500)/(9)-(625)/(9))(m//s)^2]` <br> `=750kgxx208xx208.33(m//s)^2 = 156250J`

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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.