#### Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.

Click to Chat

1800-1023-196

+91 7353221155

CART 0

• 0
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

# 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

## 2 Answers

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

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

## ASK QUESTION

Get your questions answered by the expert for free