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?

Question

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?

Philoid · Accepted 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)mv² – (1/2)mu²

⇒ W = (1/2) × 0.5 kg × ( m s^-1)² – (1/2) × 0.5 kg × 0²

⇒ W = 50 J

A body of mass 0.5 kg travels in a straight line with velocity v =a x3/2 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?

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?