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a cart moves on a smooth horizontal surface due to an external constant force F the intial mass of the the cart is M and the velocity is 0. sand falls on the cart with negligible velocity at constant rate U kg/s and sticks to the cart . find the velosity of the cart at the time T . ALSO FIND THE RATE OF INCREASE OF THE KINETIC ENERGY OF THE CART. please answer the above question in detail

a cart moves on a smooth horizontal surface due to an external constant force  F  the intial mass of the  the cart is M and the velocity is 0. sand falls on the cart with negligible velocity at constant rate U kg/s and sticks to the cart . find the velosity of the cart at the time T . ALSO FIND THE RATE OF INCREASE OF THE KINETIC ENERGY  OF THE CART.


please answer the above question  in detail


 

Grade:12

2 Answers

Pratham Ashish
17 Points
11 years ago

at any time ,

a = F/(M+m),

where m = Ut,   [dm/dt = U]

now writing acceleration as dv/dt

dv/dt = F/(M+Ut)

now integrating both sides ,we get

V = F*ln(1+Ut/M)/U ....here initial conditions are at t= 0,V=0, m = 0


now k.E , k = 0.5*(M+m)V2

rate of change of kE is dk/dt

dk/dt = 0.5*U*V2 + 0.5*(M+m)*2*V*dv/dt


dk/dt = 0.5*U*V2 + (M+m)*V*dv/dt


now just substituting the values of V , m, dV/dt


we can get the rate of change of kE


 

anandpal shekhawat
8 Points
11 years ago
hi, mass of system at any time t= M+Ut as we know that mdv/dt =Fext. + Vrel.(of variable mass w.r.t.cart)dm/dt so, we can write (M + Ut)dv/dt = F + 0*U dv = int[dt/(M/F+Ut/F] limits 0 to t so, v = ln[(M/F+Ut/F)/(M/F)] or v = ln[(M+Ut)/M] hence we can find K.E. & its rate of change

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