Q. 204.3( 26 Votes )

# A body of mass 0.5 kg travels in a straight line with velocity v =a x^{3/2} where a = 5 m^{–1/2} s^{–1}. What is the work done by the net force during its displacement from x = 0 to x = 2 m?

Answer :

Given,

Mass of the body, m = 0.5 kg

Velocity of the body is given by

v =a x^{3/2} where a = 5 m^{–1/2} s^{–1}

At x = 0, initial velocity, u = 0

At x = 2, final velocity, v = 5 (2)^{3/2} = m s^{-1}

According to work-energy theorem, work done is equal to the change in kinetic energy of the body.

Work done, W = ΔK

⇒ W = (1/2)mv^{2} – (1/2)mu^{2}

⇒ W = (1/2) × 0.5 kg × ( m s^{-1})^{2} – (1/2) × 0.5 kg × 0^{2}

⇒ W = 50 J

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