A light spring of force constant 3.55 N/m is compressed by 8.00 cm and held betw

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 0.600 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 .250 kg block .600 kg block

0.00 ___________m/s _____________m/s

0.10 ___________m/s _____________m/s

0.40 ___________m/s _____________m/s

Explanation / Answer

part A :

conservation of momentum m1v1 = m2v2

from conservation of eenrgy 0.5m1v1^2 + 0.5 m2v2^2 = 0.5 kx^2

solving for v1 and v2, we get   v1 = 2.4 v2

so

0.125 v1^2 + 0.3 V2^2 = 1.775 * 0.08* 0.08

0.125 * 2.4 v2^2 + 0.3 v2^2 = 0.0114

solving it for v2

v2 = sqrt(0.0114/1.02)

V2 = 0.106 m/s

so v1 = 0.106 * 2.4 = 0.25 m/s

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due to frictional force

0.5mm1v1^2 = u m gd1

d1 = v1^2/2ug

0.5 m2v2^2 = uk m2 g d2

d2 = v2^2/2uk g

d2 = 0.106^2 /(2 * 0.1* 9.81)

d2 = 5.73 mm

from the equation 0.5 m1v1^2 + uk gd m1 = 0.5 kx^2

v162 = 2* (0.5* 3.55* 0.08* 0.08) -(0.1* 2.5* 9.8* 0.032)/(0.25)

v1 = 0.028 m/s

similarly

0.5 m2v2^2 + uk m2 g d = 0.5 kx62

v1 =    2* (0.5* 3.55* 0.08* 0.08) -(0.1* 0.6 9.8* 0.032)/(0.6)

v1 = 0.026 m/s

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similrly for uk = 0.4

v1 = -0.16 m/s

v2 = -7.05 m/s

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