A cart loaded with sand having total mass `m_(0)=1800kg` moves on a straight horizontal road starting from rest under the action of a force of 120N. The sand spills through a small hole hole in the bottom at a rate of 0.5 kg/sec. What will be the vertically of cart after 20 min?

Answer

`120`

Explanation

In this case sand falls on th ground vertically and comes to rest. Hence horizontal velocity of ejected sand with respect to ground is zero so
`F_("ext")=d/(dt)(mv)"or" F_("ext")dt`
`=d(mv)`
Which on integration gives
`F_("ext")int_(t_(1))^(t_(2))dt=ind(mv)`
`F_("ext")=(t_(2)-t_(1))=(m_(2)v_(2)-m_(1)v_(1))`
.............i
The rate of leakage of sand
`K=(-dm//dt)` is constant which on integration gives
`int_(m_(1))^(m_(2))dm=-Kint_(t_(1))^(t_(2))dt`
`(m_(2)-m_(1))=-K(t_(2)-t_(1))`
..........ii
Here
`t_(1)=0, t_(2)=t,v_(1)=0,v_(2)=v`
From eqns i and ii
`F_("ext").t=(m_(0)-Kt)v`
`v=(F_("ext").t)/((m_(0)-Kt))`
Substituting for `F_(ext),t,K` and `m_(0)` we get
`v=(120xx20x60)/((1800-0.5xx20xx60))=120ms^(-1)`