i also need to find t but the v im coming up with is just wrong Box A, which wei

Question

i also need to find t but the v im coming up with is just wrong

Box A, which weighs 45 lb. is released from rest in the position shown in the top figure, where h = 4 ft. The box slides down the fixed incline with negligible friction and it lands on a cart B, which is initially at rest and weighs 10 lb. When A reaches the bottom of the incline, the velocity of A is completely horizontal and A starts sliding on the cart. The coefficient of kinetic friction between A and B is mu k = 0.8. Determine the common speed of A and B after A stops sliding relative to B and the time it takes A to stop sliding relative to B .

Explanation / Answer

Applying conservation of energy,

mgh=1/2 mv^2

v=(2gh)^0.5

v=(2*32*4)^0.5=16 ft/s

Velocity of A when it reaches the bottom of the incline= 16 ft/s

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Now, considering the cart B and A as a single system, the linear momentum of both the bodies should be conserved.

m(cart)*v(final)+ m(A) v(final) = m(cart)*v(initiall)+ m(A) v(initial)

Since, the initial velocity of cart is zero, and the final velocity of both are equal

10*v+45*v=0+45*16

v=45*16/55

So, Va=Vb=13.09 ft/s

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Now considering A,

N=mg

Fnet= (mew)*N= (mew)*mg

accelaration=(mew)*g=0.8*32=25.6 ft/s^2

v=13.09

u=16

v=u+at

13.09= 16 -25.6t

t=(16-13.09)/25.6=0.114 seconds

time taken for A to stop sliding= 0.114 seconds

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