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

Expert's answer

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'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*v**^{2}/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)*v**_{1}

where m - mass of the first cart, M-mass of the second cart

Hence**0.1*3=(0.1+0.2)*v**_{1}

**v**_{1}=*0.1*3/0.3=*1 m/s - velocity of twa carts after collision.

Then The height

Initial mechanical energy P=potencial

Total energy when the cart is down

Also

Kinetical energy

Hence

Also

Hence friction force is

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:

where m - mass of the first cart, M-mass of the second cart

Hence

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