Question

A force of $1000N$ acts on a particle parallel to its direction of motion which is horizontal. Its velocity increases from $1m/s$ to $10m/s$, when the force acts through a distance of $4m$. Calculate the mass of the particle. Given a force of $10N$ is necessary for overcoming friction.

Answer 1

Given force$F=1000N$.
Given that force of $10N$ is necessary for overcoming friction.
$F=1000-10$
$=990N$.
Given its velocity increases from $1m/s$ to $10m/s$ through distance of $4m$.
Here, Initial velocity $u=1$, Final velocity $v=10,S=4m$.
We know that $v^2-u^2=2as$.
${10}^2-1^2=2\times a\times 4$
$100-1=8a$
$99=8a$
$a=\frac{99}{8}$.
Now,
We have $a=\frac{99}{8},F=990,m=?$
We know that $F=ma$.
$990=m\frac{99}{8}$
$990\times 8=99m$
$7920=99m$
$m=\frac{7920}{99}$
$m=80kg$
Therefore, the mass of the particle $=80kg$.