Question

A force of $1000N$ on a particle parallel to its direction of motion which is horizontal.Its velocity increase from $1m{s}^{-1}$ to $10m{s}^{-1}$,when the force acts through a distance of $4m$.calculate the mass of the particle, given a force of $10$ newton is necessary for overcoming friction.

Answer 1

Given, $F=1000N,V_1=1m.V_2=10m/sd=4m{F}_r=10N$

Now,

$1000-10=ma\Rightarrow 990=ma............\left(1\right)$

Using the equation of motion:

$v^2=u^2+2as$

$\Rightarrow {10}^2=1^2+2\times a\times 4$

$\Rightarrow (100-1)=8a\Rightarrow a=\frac{99}{8}$

Put this value in equation (1) then we get,

$990=m\times \frac{99}{8}\Rightarrow m=\frac{990\times 8}{99}=80kg$