Question

A cart loaded with sand moves along a horizontal floor due to a constant force F coinciding in direction with the cart's velocity vector. In the process sand spills through a hole in the bottom with a constant rate $\mu$kg/s. Find the acceleration and velocity of the cart at the moment t,.if at the initial moment t = 0 the cart with loaded sand had the mass $m_0$ and its velocity was equal to zero. Friction is to be neglected.

• A. $v=\frac{1}{\mu }ln\left(\frac{m_0}{m_0-\mu t}\right)$ ; $a=\frac{F}{m_0-\mu t}$
• B. $v=\frac{F}{\mu }ln\left(\frac{m_0}{m_0-\mu t}\right)$ ; $a=\frac{1}{m_0-\mu t}$
• C. $v=\frac{F}{\mu }ln\left(\frac{m_0}{m_0-\mu t}\right)$ ; $a=\frac{F}{m_0-\mu t}$
• D. $Noneofthese$

Answer 1

The correct option is C $v=\frac{F}{\mu }ln\left(\frac{m_0}{m_0-\mu t}\right)$ ; $a=\frac{F}{m_0-\mu t}$

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 acquire the same velocity as that of the cart at the moment.
$v_r=0$Thus, $F_t=0(as{F}_t=v_r\frac{dm}{dt})$
and the net force will be F only.
$\therefore F_{net}=F$
or $m\left(\frac{dv}{dt}\right)F$............(i)
But here $m=m_0-\mu t\therefore (m_0-\mu t)\frac{dv}{dt}=F$
or ${\int }_0^{v}dv={\int }_0^t\frac{Fdt}{m_0-\mu t}\therefore v=\frac{F}{-\mu }{\left[ln\right(m_0-\mu t\left)\right]}_0^t$
or $v=\frac{F}{\mu }In\left(\frac{m_0}{m_0-\mu t}\right)$
From eq. (i), acceleration of the cart
$a=\frac{dv}{dt}=\frac{F}{m}$
or $a=\frac{F}{m_0-\mu t}$