A light spring of force constant 3.55 N/m is compressed by 8.00 cm and held betw
ID: 1340938 • Letter: A
Question
A light spring of force constant 3.55 N/m is compressed by 8.00 cm and held between a 0.250 kg block on the left and a 0.580 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.
uk = 0.0
uk= 0.100
NOTE: Please do not answer this question unless you are 100 percent sure of the answer, I am failing my physics class right now and I cannot afford to get this wrong on my homework. I have tried and tried to work the solution but i cant get it. Also, if you answer PLEASE explain the solution so I can understand it. Thank you!
Explanation / Answer
The force exerted by the spring on each block is in magnitude
F = kx
= 3.55N/m x 0.08 m
= 0.284 N.
This value must be compared to the static friction force exerted on
each block
assume mu_s=0.12
The lighter block to the left experiences the force of static friction
fl = sN
= sm*g
= 0.12 x 0.25 kg x 9.80 m/s2 = 0.294 N.
The block to the right experiences friction
fr = sN
= smr g
= 0.12 x 0.58 kg x 9.80 m/s2 =0.682 N.
Static friction for this block is greater that the spring force,
The lighter block will gain speed as long as the spring force is larger than the kinetic friction force:
the spring compression xf becomes of the value given by
0.1 · 0.25 kg · 9.80 m/s2 = xf · 3.55 N/m
xf = 0.012 m.
the energy of the lighter block as it moves to the maximum speed
Ki + Ui – fk d = Kf + Uf
0 + (3.55 x 0.082 ) J /2 - (0.1 x 0.25kgx 9.80)N x (0.08 - 0.012)m
= ½ (0.25kg x v2max) + ½ (3.55 N/m (0.012)2)
vmax = 0.0142 m/s
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