A light spring of force constant 4.45 N/m is compressed by 8.00 cm and held betw

Question

A light spring of force constant 4.45 N/m is compressed by 8.00 cm and held between a 0.250 kg block on the left and a 0.550 kg block on the right. Both blocks are at rest on a horizontal surface. The blocks are released simultaneously so that the spring tends to push them apart. Find the maximum velocity each block attains if the coefficient of kinetic friction between each block and the surface is the following. In each case, assume that the coefficient of static friction is greater than the coefficient of kinetic friction. Let the positive direction point to the right.

Explanation / Answer

From energy: (Uk=Kinetic energy, Us=Spring potential energy)
Uk1 + Uk2 = Us
1/2(m1*v1^2)+1/2(m2*v2^2 )= 1/2k*x^2
so m1*v1^2+m2*v2^2 = k*x^2
plug in known values:
.25*v1^2+.55*v2^2 = 4.45N/m(.08m)

From momentum: (p=momentum)
we know that p1=p2
m1*v1 = m2*v2
so v1 = (m2*v2)/m1 or with values v1 = (.55*v2)/.25 => v1 = 2.2*v2
when sqaured v1^2 = 4.84*v2^2

now we use value of v1^2 in energy equation ::
.25*4.84*v2^2+.55*v2^2 = 4.45N/m(.08m)

v2 = .45 m/sec
v1 = .99 m/sec

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