Question

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A body of mass 10 kg slows down fror
10 m/s to 5 m/s then work done on it
is​

Answer 1

Answer:

work done is equal to change in kinetic energy

= K f - K i

= (1/2×m×v^2)-(1/2×m×u^2)

= (1/2×10×5^2)-(1/2×10×10^2)

= (125)-(500)

= -375 N

negative sign indicates opposite direction

Answer 2

Given : A body of mass 10 kg slows down from 10 m/s to 5 m/s

To Find : work done

Solution:

m = 10 kg

u = 10  m/s

v= 5 m/s

Initial Kinetic energy = (1/2)mu²

= (1/2)* 10 * 10²

= 500  J

Final Kinetic energy = (1/2)mv²

= (1/2)* 10 * 5²

= 125  J

Work done = 125 - 500 = - 375 J

work done on body = - 375J

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