What is the work done to increase the velocity of a car from 30 km/h to 60 km/h if the mass of the car is 1500 kg class 9?

Last updated at June 8, 2019 by Teachoo

Example 11.4 What is the work to be done to increase the velocity of a car from 30 km h–1 to 60 km h–1 if the mass of the car is 1500 kg? Mass of the car = m = 1500 kg Initial velocity = u = 30 km/h = 30 × 5/18 m/s = (5 × 5)/3 = 25/3 m/s Final velocity = v = 60 km/h = 60 × 5/18 m/s = (10 × 5)/3 = 50/3 m/s Conversion from km/h to m/s 1 km = 1000 m 1 hour = 60 minutes = 3600 s ∴ (1 𝑘𝑚)/ℎ = (1000 𝑚)/(3600 𝑠) ∴ 1 km/h = 5/18 m/s We know that Work done = Change in kinetic energy Finding initial and final kinetic energy Initial Kinetic energy = 1/2 mu2 = 1/2 × (1500) × (25/3) (25/3) = (750 × 625)/9 = (250 × 625)/3 = (156250 J)/3 Final Kinetic energy = 1/2 mv2 = 1/2 × (1500) × (50/3) (50/3) = (750 × 2500)/9 = (250 × 2500)/3 = (625000 J)/3 Work done = Change in kinetic energy = Final kinetic energy − Initial kinetic energy = 625000/3 − 156250/3 = 468750/3 = 156250 J Work done is 156250 J

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.