Question

The rate of doing work by force acting on a particle
moving along x-axis depends on position x of
particle and is equal to 2x. The velocity of particle
is given by expression

Answer 1

Answer:

Rate of doing work is power so according to the given condition

P = 2x

so Fv = 2x

or m(dv/dt)v = 2x

or m(vdv/dx)v = 2x

or mv2dv = 2xdx

now integrate.

answer :(3X2/m)1/3

Answer 2

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

The rate of doing work (Power) depends on position of the particle.

$\huge{\bold{\underline{Explanation:-}}}$

As mentioned in question,

$\large{\bold{ \frac{dw}{dt} = 2x}}$

Substituting the work done formula,

$\large{ \frac{d(F.s)}{dt} = 2x}$

Here F is constant and cannot be differentiated.

$\large{ F. \frac{ds}{dt} = 2x}$

ds/dt = velocity.

Substituting the value

$\large{\bold{F.v = 2x}}$

From Newton's Second law we know that F = ma.

Substituting the Force value

$\large{ma.v = 2x}$

Now, acceleration formula,

$\large{m \times \frac{dv}{dt} \times v = 2x}$

Now,

$\large{m \times (\frac{dv}{dt} \times \frac{dx}{dx})\times v = 2x}$

Rearranging,

$\large{m \times ( \frac{dx}{dt} \times \frac{dv}{dx}) \times v = 2x}$

As we know dx/dt is Velocity.

$\large{ m \times v \times v \times \frac{dv}{dx} = 2x}$

$\large{ m \times v^2 \times \frac{dv}{dx} = 2x}$

Now,

$\large{m \times v^2 dv = 2x.dx}$

Integrating,

$\large{ m \displaystyle\int v^2.dv = \displaystyle\int 2x.dx}$

Here mass cannot be Integrated as it is constant.

Applying limits,

$\large{m \displaystyle\int^v_0 v^2.dv = \displaystyle\int^x_0 2x.dx}$

Integrating,

$\large{m \left| \frac{v^3}{3} \right|^v_0 = \left| \frac{2x^2}{2} \right|^x_0}$

$\large{m \left| \frac{v^3}{3} \right|^v_0 = \left| \frac{ \cancel{2}x^2}{ \cancel{2}} \right|^x_0}$

Now,

$\large{m \frac{v^3}{3} = x^2}$

$\large{ \frac{v^3}{3} = \frac{x^2}{m}}$

$\large{v^3 = \frac{3x^2}{m}}$

$\large{ v = \sqrt[3]{ \frac{3 {x}^{2} }{m} }}$

$\huge{\boxed{\boxed{v = \left[ \frac{3x^2}{m} \right]^{(1/3)}}}}$