# Answer to Question #3212 in Mechanics | Relativity for Bella

Question #3212
A 100g cart slides down a ramp angled 30 degrees above the horizontal and arrives at the bottom with a speed 3m/s. How much mechanical energy is lost as the cart slides down the ramp. find the strength of the frictional force acting on the cart. If the cart collides with a second cart, of mass 200g, at the bottom of the ramp and sticks, find the velocity of the two carts
1
2011-06-23T12:11:27-0400
First, to solve the problem you should know height from which cart slideds or length of the way it covers sliding down otherwise you wouldn&#039;t be able to find the initial value of mechanical energy. So we denote the length of the way down as S.
Then The height H=S*sin(30)=S/2
Initial mechanical energy P=potencial
energy = m*g*H=m*g*S/2 = 0.1*9.81*S/2 = 0.49*S
Total energy when the cart is down E = K+A(F) where A(F)- work of friction force
Also P = E = K+A(F) due to conservation of energy law
Kinetical energy K=m*v2/2=0.1*9/2=0.45
Hence A(F) = 0.49S-0.45 - this value denotes how much mechanical energy is lost
Also A(F) = F*S
Hence friction force is F=(0.49S-0.45)/S

If the cart collides with a second cart, of mass 200g, at the bottom of the ramp and sticks then due to the Law of impuls conservation: m*V=(m+M)*v1
where m - mass of the first cart, M-mass of the second cart
Hence 0.1*3=(0.1+0.2)*v1
v1=*0.1*3/0.3=*1 m/s - velocity of twa carts after collision.

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