Question

A particle of mass 500 gm is moving in a straight line with velocity v = bx^5/2. The work done by the net force during its displacement from x = 0 to x = 4 m is : (Take b = 0.25 m^-3/2 s^-1).

Options

• 2J

• 4J

• 8J

• 16J

Answer 1

16J

Explanation:

Given: Velocity as a function of position as v = bx^5/2    ...(1)

By differentiating the above equation with respect to time

`"a" = "dv"/"dt" = "d"/"dt"` (bx^5/2)  = `5/2`bx^3/2

⇒ a = `5/2` bx^3/2 × v        ...[since, v = `dx/dt`]

⇒ a = `5/2` bx^3/2 × b × x^5/2 

⇒  a = `5/2`bx⁴

Mass of body = 500 gm = 0.5 kg

Force acting on body = m × a         ...(by Newton’s II law of motion)

F = 0.5 × `5/2`bx² x⁴

= `5/4`b²x⁴

By definition of work done ⇒ W = `int_(x_1)^(x_2)` Fdx

⇒ W = `int_(x=0)^(x=4) 5/4b^2x^4dx`

⇒ `(5b^2)/4 int_(x=0)^(x=4) x^4dx`

⇒ W = `5/4 × (1/4)^2 [x^5/5]_(x=0)^(x=4)`

= `5/4 xx 1/16 xx(4^5 - 0)/5 = 1024/64` 

⇒ W = 16J