The speed of a vehicle of mass $500 k g$ increases from $30 m s^{- 1}$ to $70 m s^{- 1}$ . Calculate the work.

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Solution

Initial Kinetic energy , $K_{1} = \frac{1}{2} \times 500 \times 30^{2} = 250 \times 900 = 225000 J$

Final Kinetic energy , $K_{2} = \frac{1}{2} \times 500 \times 70^{2} = 250 \times 4900 = 1225000 J$

Work = Change in Kinetic energy , $= K_{2} - K_{1} = 1225000 - 225000 = 1000000 = 10^{6} J$

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The speed of a vehicle of mass 500kg increases from 30ms−1 to 70ms−1. Calculate the work.

Initial Kinetic energy ,K1=12×500×302=250×900=225000 JFinal Kinetic energy ,K2=12×500×702=250×4900=1225000 JWork = Change in Kinetic energy , =K2−K1=1225000−225000=1000000=106 J

The speed of a vehicle of mass $500 k g$ increases from $30 m s^{- 1}$ to $70 m s^{- 1}$ . Calculate the work.

Initial Kinetic energy , $K_{1} = \frac{1}{2} \times 500 \times 30^{2} = 250 \times 900 = 225000 J$

Final Kinetic energy , $K_{2} = \frac{1}{2} \times 500 \times 70^{2} = 250 \times 4900 = 1225000 J$

Work = Change in Kinetic energy , $= K_{2} - K_{1} = 1225000 - 225000 = 1000000 = 10^{6} J$