Question

A body of mass 0.5 kg travels in a straight line with velocity v = ax^3/2 where a = 5 m^–1/2s^–1. The change in kinetic energy during its displacement from x = 0 to x = 2 m is ______.

Options

• 1.5 J

• 50 J

• 10 J

• 100 J

Answer 1

A body of mass 0.5 kg travels in a straight line with velocity v = ax^3/2 where a = 5 m^–1/2s^–1. The change in kinetic energy during its displacement from x = 0 to x = 2 m is 50 J.

Explanation:

Given m = 0.5 kg, v = kx^3/2 where, k = 5 m^–1/2s^–1

Acceleration, a = `"dv"/"dt"="dv"/("d"x)("d"x)/"dt"="v""dv"/("d"x)(because "v" = "dx"/"dt")`

As v² = k²x³

Differentiating both sides with respect to x, we get

2v`"dv"/("d"x)` = 3k²x³

Acceleration, a = `3/2`k²x³

Force, F = Mass × Acceleration = `3/2`k²x³

 

Work done, W = `int"Fd"x=int_0^2 3/2"mk"^2x^2"d"x`

W = `3/2"mk"^2[x^3/3]_0^2`

= `3/6xx0.5xx5^2xx[2^3-0]`

= 50 J.