Question

A cart loaded with sand moves along a horizontal floor due to constant force $F=10\text{N}$ acting in the direction of the cart's velocity vector. In the process, sand spills through a hole at the bottom at a constant rate, $\mu =5\text{kg/s}$ . Find the velocity of the cart at the time $t=5\text{sec}$, if at the initial moment $t=0$, the cart loaded with sand has mass $m_0=100\text{kg}$ and its velocity was equal to zero. Friction is to be neglected. [Take $\ln \left(\frac{4}{3}\right)=0.288$ and $g=10{\text{m/s}}^2$]

• A. $1.428\text{m/s}$
• B. $0.823\text{m/s}$
• C. $1.1\text{m/s}$
• D. $0.576\text{m/s}$

Answer 1

The correct option is D $0.576\text{m/s}$

Given,
Initial mass of the cart + sand $m_0=100\text{kg}$
Rate at which sand spills from the hole $\mu =5\text{kg/s}$

In this problem, the sand spills through a hole in the bottom of the cart. Hence, the relative velocity of the sand $v_r$ will be zero because it will have the same horizontal velocity as the cart at the moment of spilling.

Thus, $F_t=0[\text{as}{F}_t=v_r\left(\frac{-dm}{dt}\right)]$
and the net force will be $F$ only.
i.e $F_{net}=F$
or $m\left(\frac{dv}{dt}\right)=F$ ...(i)
But here, $m=m_0-\mu t$
$\Rightarrow (m_0-\mu t)\frac{dv}{dt}=F$


Integating both sides,
${\int }_0^{v}dv={\int }_0^t\frac{F}{(m_0-\mu t)}dt$
$\Rightarrow v=\frac{F}{-\mu }\left[\ln \right(m_0-\mu t){]}_0^t$
$\Rightarrow v=\frac{F}{\mu }\ln \left(\frac{m_0}{m_0-\mu t}\right)$
At $t=5\text{s}$,
$v=\frac{10}{5}\ln \left(\frac{100}{100-5\times 5}\right)=2\ln \left(4/3\right)$
$\therefore v=0.576\text{m/s}$