Question

A body is moving with velocity v =lnx m/s where x is position .the net force on body is zero at

Answer 1

Newton's Second Law gives the definition of Force.

The rate of change of momentum with respect to time is Force

In Mathematical Form,

$\displaystyle \vec{F}=\frac{d\vec{p}}{dt}$

where $\vec{p}$ is momentum.

Since we write $\vec{p}=m\vec{v}$, we can write Force as:

$\displaystyle F = m\frac{d\vec{v}}{dt}$

We are given velocity here.

$v=\ln x \, \, m/s$

We need to differentiate it with respect to time. But velocity is given as a function of position. So we differentiate it as follows:

$\displaystyle v = \ln x \\ \\ \\ \implies \frac{dv}{dt} = \frac{d}{dx} \, (\ln x) \times \frac{dx}{dt} \\ \\ \\ \implies \frac{dv}{dt} = \frac{1}{x} \times v \\ \\ \\ \implies \frac{dv}{dt} = \frac{1}{x} \times \ln x \\ \\ \\ \implies \frac{dv}{dt} = \frac{\ln x}{x}$

Now, we have to find where force is zero.

$\displaystyle F=0 \\ \\ \\ \implies m\frac{dv}{dt}=0 \\ \\ \\ \implies m \times \frac{\ln x}{x}=0 \\ \\ \\ \implies \frac{\ln x}{x}=0 \\ \\ \\ \implies \ln x=0 \\ \\ \\ \implies \boxed{\bold{x=1 \, \, m}}$

Thus, Force becomes zero at $\bold{x=1 \, \, m}$

Answer 2

a=ma

a=mdv/dt

a=md(lnx)/dt