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ICSE Class 10 Physics Question 19 of 23

Work, Energy and Power — Question 6

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Question

Question 6

A body of mass m is moving with a uniform velocity u. A force is applied on the body due to which its velocity increases from u to v. How much work is being done by the force?

Answer

Mass = m

Change in velocity from u to v, so acceleration acts (a)

Force = f

Work done by the force = force × displacement

From equation of motion we know,

v2=u2+2aS∴S=v2−u22av^2 = u^2 + 2aS \\[0.5em] \therefore S = \dfrac{v^2 -u^2}{2a}

Substituting it in the formula for work done:

W=m×a×(v2−u22a)W=(12×m×v2)−(12×m×u2)W=(12×m×v2)−(12×m×u2)W=12m(v2−u2)W=KEf−KEiW = m \times a \times (\dfrac{v^2-u^2}{2a}) \\[0.5em] W = (\dfrac{1}{2} \times m \times v^2) - (\dfrac{1}{2} \times m \times u^2) \\[0.5em] W = (\dfrac{1}{2} \times m \times v^2) - (\dfrac{1}{2} \times m \times u^2) \\[0.5em] W = \dfrac{1}{2} m (v^2 - u^2) \\[0.5em] W = KE_f - KE_i