Question

The horizontal surface on which the block slides is frictionless. The speed vi of the block of mass m= 2.0 kg before it touches the spring is 10 m/s. After the spring is compressed by 0.20 m, the block starts to move back to the right. What is the velocity vf of the block in m/s when the spring is compressed by 0.15m ?

Answer #1

Initial energy of the block = 0.5*mass*(speed before it touches the spring)^2 = 0.5*2*vi^2

So, Initial energy of the block = 0.5*2*10^2 = 100J

Total energy of the block and spring system when the spring is compressed by 0.20 m and the block starts to move back to the right = 0.5*2*0^2+0.5*k*0.2^2 = 0.5*k*0.2^2

where k = spring constant

Total energy is constant at any time

So, 0.5*k*0.2^2= 100J

So, spring constant = k = 5000N/m

Final energy of the block and spring system when the spring is compressed by 0.15m = 0.5*mass*(speed of the block in m/s when the spring is compressed by 0.15m)^2 + 0.5*k*0.15^2

So, 0.5*2*vf^2 + 0.5*5000*0.15^2 = 100J

So, vf = 43.75^1/2 = 6.6 m/s

So, the velocity vf of the block in m/s when the spring is compressed by 0.15m is 6.6 m/s

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