A light spring of force constant 3.50 N/m is compressed by 8.00 cm and held betw

Question

A light spring of force constant 3.50 N/m is compressed by 8.00 cm and held between a 0.250 kg block on the left and a 0.510 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 0.000 0.100 0.470 k 0.250 kg block 0.510 kg block m/s m/s m/s m/s m/s m/s

Explanation / Answer

Given

let m1 = 0.25 kg , m2 = 0.510 kg on right and left side of the spring  

k = 3.50 N , x = 0.08 m

the maximum speed for the mass will be at the equilibrium so  

by conservation of eneryg  

elastic potential energy - frictional force *x = change in kinetic energy

for m1

with  

mue_k = 0 , W_f = 0

0.5*k*x^2- W_f = 0.5*m1*v^2

0.5*3.5*0.08^2 - 0 = 0.5*0.25*v^2

v = 0.299 m/s = 0.3 m/s

with  

mue_k = 0.1 , W_f = mue_k*m1*g = 0.1*0.25*9.8*0.08 = 0.0196 J

0.5*k*x^2- W_f = 0.5*m1*v^2

0.5*3.5*0.08^2 - 0.0196 = 0.5*0.25*v^2

v = 0.259 m/s

with  

mue_k = 0.47 , W_f = mue_k*m1*g = 0.47*0.25*9.8*0.08 = 0.09212 J

0.5*k*x^2- W_f = 0.5*m1*v^2

0.5*3.5*0.08^2 - 0.09212 = 0.5*0.25*v^2

v = 0.2011 m/s

for m2

with  

mue_k = 0 , W_f = 0

0.5*k*x^2- W_f = 0.5*m1*v^2

0.5*3.5*0.08^2 - 0 = 0.5*0.510*v^2

v = 0.2096 m/s  

with  

mue_k = 0.1 , W_f = mue_k*m1*g = 0.1*0.510*9.8*0.08 = 0.039984 J

0.5*k*x^2- W_f = 0.5*m1*v^2

(0.5*3.5*0.08^2 - 0.039984) = 0.5*0.51*v^2

v = 0.1129 m/s

with  

mue_k = 0.47 , W_f = mue_k*m1*g = 0.47*0.51*9.8*0.08 = 0.1879248 J

0.5*k*x^2 - W_f = 0.5*m1*v^2

(0.5*3.5*0.08^2 - 0.1879248) = 0.5*0.51*v^2

v = 0.8324 m/s

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