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.