You have a spring compressed 0.20m that has a spring constant of 225N/m. A 2kg b

Question

You have a spring compressed 0.20m that has a spring constant of 225N/m. A 2kg block at rest is placed against the compressed spring and is on a frictionless surface. Then the spring is released, pushing the block forward; it travels 20cm until it reaches a rough surface. (coefficient of kinetic friction of rough surface is 0.20)

After moving 10cm over the rough surface, the block has an inelastic collision with another block of mass 3.0kg. (They stick together)

1.What is the speed of the 2.0 kg block when it leaves the spring?

2.How fast is the 2.0 kg block moving when it collides with the 3.0 kg block?

3.How far will the two blocks travel before coming to rest?

Explanation / Answer

1)

By energy conservation, the intitial potential energy possessed by the spring is converted to kinetic energy of the block.

So, Kinetic energy(K.E) = potential enrgy(P.E)

So, 0.5*mv^2 = 0.5*k*x^2

So, v = sqrt(kx^2/m) = sqrt(225*0.2^2/2) = 2.12 m/s <----------answer

b)

by work energy principle,

initial K.E of the block - work done due to friction = final K.E of the block when it hets the other block

So, 0.5*mv^2 - umgd = 0.5*mv'^2

so, 0.5*v^2 - ugd = 0.5*v'^2

So, 0.5*2.12^2 - 0.2*9.8*0.1 = 0.5*v'^2

So, v' = 2.03 m/s <-----------answer

C)

conserving momentum,

m*v' =(M+m)*V

So, V = mv'/(m+M) = 2*2.03/(2+3) = 0.812 m/s

So, by conservation of energy,

0.5*(m+M)*V^2 = u(m+M)*g*d'

So, d' = 0.5*(m+M)*V^2/(m+M)*u*g)

= 0.5*V^2/ug = 0.5*(0.812)^2/(0.2*9.8)

= 0.168 m <------------answer

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