Question

A body of mass $m$ rests on a horizontal plane with a friction coefficient $\mu$. At $t=0$, a horizontal force is applied ($F=at$) where $a$ is a constant vector. Find the distance traversed in first $t$ sec.

• A. $\frac{a}{6m}{(t-\frac{\mu mg}{a})}^3$
• B. $\frac{a}{6m}{t}^3$
• C. $\frac{a}{6m}{t}^3-\frac{\mu g{t}^2}{2}$
• D. $\frac{a}{6m}{t}^3-\frac{\mu g{t}^2}{3}$

Answer 1

The correct option is A $\frac{a}{6m}{(t-\frac{\mu mg}{a})}^3$

Let at time $t_0$ the body begins to move. so, $a{t}_0=\mu mg$
or, $t_0=\frac{\mu mg}{a}$
now, $m\frac{dv}{dt}=a(t-t_0)$
put $t-t_0=T$
or, $dv=\frac{a}{m}TdT$
integrating both sides,$v=\frac{a{T}^2}{2m}$
as $v=\frac{dx}{dt}$, so, $x=\int \frac{a{T}^2}{2m}dT=\frac{a{T}^3}{6m}$
distance $x=\frac{a}{6m}(t-\frac{\mu mg}{a}{)}^3$