Chapter - 4 Work and Energy

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Q
Work and Energy

Question:

Calculate the work required to be done to stop a car of 1500 kg moving at a velocity of 60 km/h?

Answer:

Given: Initial velocity, u = 60 kmh^{-1}

u = 60 	imes frac{5}{18} ms^{-1}

u = frac{50}{3} ms^{-1}

Final velocity, v = 0 ms^{-1} (because car will stop)

Mass of the body, m = 1500 kg

We know, kinetic energy of a body is

K.E. = frac{1}{2}mv^2

The initial kinetic energy of the body, 

E_i = frac{1}{2}mu^2

E_i = frac{1}{2} 	imes 1500 	imes (frac{50}{3})^2

E_i = 750 	imes frac{2500}{9}

E_i = 208333.3 J

The final kinetic energy of the body, 

E_f = frac{1}{2}mv^2

E_f = frac{1}{2} 	imes 20 	imes (0)^2

E_f = 0

Work done = Change in kinetic energy

Work done = Final K.E. - Initial K.E.

Work done = 0 J - 208333.3 J

Work done = -208333.3 J

Here, the negative sign shows that the force is acting in the opposite direction of the motion of the object.


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