Question

A flatcar of mass $m_0=20\text{kg}$ is moving to the right due to a constant horizontal force $F=10\text{N}$. Sand spills on the flatcar from a stationary hopper. The velocity of loading is constant and equal to $\mu =1\text{kg/s}$.



Find the acceleration of the flatcar after $t=20\text{s}$.

• A. $\frac{1}{8}{\text{m/s}}^2$
• B. $\frac{1}{40}{\text{m/s}}^2$
• C. $0$
• D. $5{\text{m/s}}^2$

Answer 1

The correct option is A $\frac{1}{8}{\text{m/s}}^2$

Given, $F=10\text{N}$
For a variable mass system where mass is added at a rate $\mu$,
$m\cdot \frac{dv}{dt}=F_{ext}+\mu u$
where $v$ = velocity of car, $m_0$ = mass of car $=20\text{kg}$ and $\mu =1\text{kg/s}$
Here, $m=m_0+\mu t$ after $t$ seconds
and relative velocity of mass added $u=-v$
(in horizontal direction, sand has zero velocity)
$\Rightarrow (m_0+\mu t)\frac{dv}{dt}=F-\mu v$

Rearranging the terms,
$\Rightarrow m_{0}dv+\mu tdv=Fdt-\mu vdt$
$\Rightarrow m_{0}dv+\mu (tdv+vdt)=Fdt$
Intergrating $v$ from $0$ to $v$ and $t$ from $0$ to $t$:
$\Rightarrow \int m_{0}dv+\int \mu (tdv+vdt)=\int Fdt$
$\Rightarrow m_{0}v+\int \mu d\left(vt\right)=Ft$
$\Rightarrow m_{0}v+\mu vt=Ft$
or $v=\frac{Ft}{(m_0+\mu t)}=\frac{10t}{20+t}$

Now, differentiating both sides
$\frac{dv}{dt}=\frac{(m_0+\mu t)F-Ft\left(\mu \right)}{(m_0+\mu t{)}^2}$
$\Rightarrow a=\frac{m_{0}F}{m_0^2{(1+\frac{\mu t}{m_0})}^2}=\frac{10}{m_0{(1+\frac{\mu t}{m_0})}^2}$
$\Rightarrow a=\frac{10}{20{(1+\frac{1\times 20}{20})}^2}$
$\therefore a=\frac{1}{8}{\text{m/s}}^2$ after $t=20\text{s}$