Question

A flat car o mass $m_0$ starts moving to the right due to a constant horizontal force F. Sand spills on the flat car from a stationery hopper. The rate of loading is constant and equal to $\mu$ kg/s. find the time dependence of the acceleration of the flat car in the process of loading. The friction is negligibly small.

• A. $\frac{F{m}_0}{(m_0+\mu t{)}^2}$
• B. $\frac{F{m}_0}{(m_0+\mu t)}$
• C. $\frac{2F{m}_0}{(m_0+\mu t{)}^2}$
• D. $\frac{F{m}_0}{2(m_0+\mu t{)}^2}$

Answer 1

The correct option is A $\frac{F{m}_0}{(m_0+\mu t{)}^2}$

Answer is A.
Initial velocity of the flat car is zero. Let v be its velocity at time t and m its mass at that instant. Then
At $t=0,v=0$ and $m=m_o$ at$t=t,v=v$ and $m=m_o$
Here, $v_r=v\left(backwards\right)$.
$\frac{dm}{dt}=\mu$
Therefore, $F_t=v_r\frac{dm}{dt}=\mu v\left(backwards\right)$
Net force on the flat car at time t is $f_{net}=F-F_t$
or $m\frac{dv}{dt}=F-\mu v...\left(i\right)$
or $(m_0+\mu t)\frac{dv}{dt}=F-\mu vor{\int }_0^v\frac{dv}{F-\mu v}={\int }_0^t\frac{dt}{m_0+\mu t}$
Therefore, $-\frac{1}{\mu }\left[ln\right(F-\mu v){]}_0^v=\frac{1}{\mu }\left[ln\right(m_0+\mu t){]}_0^t$
$\Rightarrow ln\left(\frac{F}{F-\mu v}\right)=ln\left(\frac{m_0+\mu t}{m_0}\right)$
So, $\frac{F}{F-\mu v}=\frac{m_0+\mu t}{m_0}orv=\frac{Ft}{m_0+\mu t}$
from Eq.(i), $\frac{dv}{dt}$ = acceleration of flat car at time t
or $=\frac{F-\mu v}{m}a=\left(\frac{F-\frac{F\mu t}{m_0+\mu t}}{m_0+\mu t}\right)ora=\frac{F{m}_0}{(m_0+\mu t{)}^2}$.
Hence, the time dependence of the acceleration of the flat car in the process of loading is $\frac{F{m}_0}{(m_0+\mu t{)}^2}$.