Question

A freight car of mass M contains a mass of sand m. At t = 0 a constant horizontal force F is applied in the direction of rolling and at the same time a port in the bottom is opened to let the sand flow out at constant rate dm/dt. Find the speed of the freight car when all the sand is gone. Assume the freight car is at rest at t = 0.

Answer 1

Answer:

Amount of linear movement

Explanation:

Our system is defined by the rate of change in mass that

leaves the car $\Delta m_ {s}$ , this happens during a time interval

$[t, t + \Delta t]$, in addition to freight car and sand at time t.

In this way we need to define the two states:

State 1,

consider $t, m_ {c} (t) + \Delta m_ {s}$ and V.

State 2,

consider $t + \Delta t, m_ {c} (t), V + V \Delta V$

In this state is the mass of sand output, which

is composed of

$\Delta m_{s}, V + \Delta V$

In this way we define the Linear movement in x, like this:

$p_ {x} (t) = (\Delta m_ {s} + m_ {c} (t)) v$

$p_ {x} (t+\Delta t) = (\Delta m_ {s} + m_ {c} (t)) (v + \Delta v)$

$m_ {c} (t) = m_ {c, 0} - bt = m_ {c} + m_ {s} -bt$

In this way we proceed to obtain the Force

$F =\lim_{\Delta t \rightarrow 0} \frac {p_x (t + \Delta t) -p_ {x} (t)} {\Delta t}$

$F = lim_{\Delta t \rightarrow 0} m_ {c} (t) \frac {\Delta v} {\Delta t} + lim_{\Delta t \rightarrow 0} m_ {s} (t) \frac {\Delta v} {\Delta T}$

Since the mass of the second term becomes 0, the same term is eliminated, thus,

$F = m_ {c} (t) \frac {dv} {dt}$

$\int\limit ^ {v (t)} _ {v = 0} dv = \int\limit^t_0 \frac{Fdt} {m_ {c} + m_ {s} -bt}$

$V (t) = - \frac {F} {b} ln (\frac {m_c + m_s-bt} {m_c + m_s})$