A light spring of force constant 3.85 N/m is compressed by 8.00 cm and held betw

Question

A light spring of force constant 3.85 N/m is compressed by 8.00 cm and held between a 0.250 kg block on the left and a 0.460 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.

0.460 kg block

0.00 m/s ? ?

0.100 m/s ? ?

0.461 m/s ?(It's 0 show why) ?(also 0 sho why)

Explanation / Answer

you must use conservation of energy and conservation of momentum:

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+.46*v2^2 = 3.85N/m(.08m) -----(eq 1)

From momentum: (p=momentum)
we know that p1=p2
m1*v1 = m2*v2
so v1 = (m2*v2)/m1 or with values v1 = (.46*v2)/.25 => v1 =1.84*v2
when sqaured v1^2 = 3.38*v2^2

plug into the energy equation for v1^2 to find v2. Once found, plug v2 into the momentum to find v1.

.25*3.38*v2^2 +.46*v2^2 = 3.85N/m(.08m)

V2=.46m/s

V2=.83m/s

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