Question

A particle of mass m starts moving from origin along x axis and its velocity varies with position (x) as v =kx 1/2 . the work done by force acting on it during first 't' seconds is 1) (mk^4t^2)/2 2) (mk^2t)/2 3) (mk^4t^2)/8

Answer 1

mass = m.     velocity v(x) = k √x

Acceleration a (t) = dv/dt = dv/dx * dx/dt
                            = k / (2√x) * v
                            = k² / 2 

Force F(t) = m k² /2

Displacement = x(t)   as the particle moves along x axis.
    v(x) = dx/dt = k√x
=>  dx/√x = k dt
Integrating on both sides:
=>   2 √x = k t + c
=>   At  t = 0,  x = 0        and  Hence,  c= 0
=>    x = k² t² / 4
$Work\ done\ W(t)=\int \limits_{x=0}^{x} {F(x)} \, dx\\\\ =\int \limits_{0}^{x} {\frac{mk^2}{2}} \, dx \\\\=\frac{1}{2}mk^2x\\\\=\frac{1}{8}mk^4t^2$

Option (3)

Answer 2

mass = m.     velocity v(x) = k √x

Acceleration a (t) = dv/dt = dv/dx * dx/dt

                            = k / (2√x) * v

                            = k² / 2 

Force F(t) = m k² /2

Displacement = x(t)   as the particle moves along x axis.

    v(x) = dx/dt = k√x

=>  dx/√x = k dt

Integrating on both sides:

=>   2 √x = k t + c

=>   At  t = 0,  x = 0        and  Hence,  c= 0

=>    x = k² t² / 4