Question

A particle of mass $\frac{10}{7}$ kg is moving in the positive direction of x. Its initial position is x = 0 and initial velocity is 1 m/s. Find the velocity in m/s at x = 10 m, if power at x=0 is 2 watts

Answer 1

Area under P - x graph $=\int Pdx$
$=\int \left(\frac{mdv}{dt}\right)vdx$
$={\int }_1^{v}m{v}^{2}dv={\left[\frac{m{v}^3}{3}\right]}_1^v$
$=\frac{10}{7\times 3}(v^3-1)$ ...(1)
From graph area,
$=\frac{1}{2}\times (2+4)\times 10=30$ ...(2)
From (1) and (2),
$\frac{10}{7\times 3}(v^3-1)=30$
$=v=4m/s$

Why this question? Tip: Manupulations in the formula are done in a way so as to express it in variables, we have information about- $P=F.v=ma.v$, but we write $a$ as $vdv/dx$ since we have information about v and x)