Question

A cart loaded with sand moves along the horizontal floor due to a constant force \[F\] coinciding in direction with the cart’s velocity vector. In the process, the sand spills through a hole in the bottom with constant rate \[\mu \,{\text{kg}}{{\text{s}}^{{\text{ - 1}}}}\]. Find the acceleration and the velocity of the cart at the moment \[t\], if the initial moment \[t = 0\,{\text{s}}\] the cart with loaded sand has mass \[{m_o}\] and its velocity was equal to zero. Friction is to be neglected.

Answer 1

Hint: Find how the mass of the cart will change with time when the sand starts to spill through the hole. Use the formula for force to find out the acceleration and velocity of the cart.

Complete Step by step answer:
Given, mass of the cart initially is \[{m_0}\]
Force with which the cart is moving along horizontal floor is \[F\]
Rate at which the sand spills through the hole is \[\mu \,{\text{kg}}{{\text{s}}^{{\text{ - 1}}}}\]
Let \[v\] be the velocity of the cart.
The rate of decrease of the mass of cart is,
\[\dfrac{{dm}}{{dt}} = - \mu \] (negative sign as the mass is decreasing)
\[ \Rightarrow dm = - \mu dt\]
Integrating we have,
\[
\int\limits_{{m_0}}^m {dm} = - \mu \int\limits_0^t {dt} \\
   \Rightarrow m - {m_0} = - \mu t \\
   \Rightarrow m = {m_0} - \mu t \\
\]
The sand spills through the bottom, so the relative velocity of the sand \[{v_r}\] will be zero because it will acquire the same velocity as that of the cart at any moment.
Therefore, relative force is also zero.
Now, the net force is, \[{F_{net}} = F\]
We have the formula for force as,
\[F = m\dfrac{{dv}}{{dt}}\]
Putting the value of \[m\]in the above equation we get,
\[
F = \left( {{m_0} - \mu t} \right)\dfrac{{dv}}{{dt}} \\
   \Rightarrow \dfrac{{dv}}{{dt}} = \dfrac{F}{{\left( {{m_0} - \mu t} \right)}} \\
   \Rightarrow dv = \dfrac{F}{{\left( {{m_0} - \mu t} \right)}}dt \\
\]
Integrating the above equation we get
\[\int\limits_0^v {dv} = \int\limits_0^t {\dfrac{F}{{\left( {{m_0} - \mu t} \right)}}dt} \]
\[\Rightarrow v = \dfrac{F}{{ - \mu }}\left[ {\ln \left( {{m_0} - \mu t} \right)} \right]_0^t \\
   \Rightarrow v = \dfrac{F}{{ - \mu }}\left[ {\ln \left( {{m_0} - \mu t} \right) - \ln \left( {{m_0}} \right)} \right] \\
   \Rightarrow v = \dfrac{F}{\mu }\ln \left[ {\dfrac{{{m_0}}}{{{m_0} - \mu t}}} \right] \\
\]
The acceleration of the cart can be written as,
\[a = \dfrac{F}{m}\]
Putting the value of \[m\] we get,
\[a = \dfrac{F}{{{m_0} - \mu t}}\]

Therefore, velocity of the cart is \[v = \dfrac{F}{\mu }\ln \left[ {\dfrac{{{m_0}}}{{{m_0} - \mu t}}} \right]\] and acceleration of the cart is \[a = \dfrac{F}{{{m_0} - \mu t}}\].

Note: In this question the frictional forces are neglected but in questions where the frictional forces are considered then at first we should make a neat diagram carefully assigning all the forces acting on the system and then proceed for the calculations.