In the figure below m 1 = 1 kg and m2 = 2 kg. These two masses are pushed togeth
ID: 1726862 • Letter: I
Question
In the figure below m1= 1 kg and m2 = 2 kg. These two masses are pushed together on a horizontalsurface to an initial separation of xi = 5 cm. The spring that is attached to the massm1 has a spring constant of k = 3750 N/m and an equilibrium length of 7 cm. Thus, the springis initially compressed between the two masses. When the masses arereleased, they move apart from each other, but eventually come to arest because of the friction between the masses and the horizontalsurface. The coefficient of kinetic friction between the masses andthe surface is k = 0.051. What is the final separation distancexf?Explanation / Answer
m1 = 1kg, m2 = 2 kg.xi = 5 cm.k = 3750 N/m, initialcompression x = 2 cm, = 0.051. Find thefinal separation distance xf. kx2/2 = m1*v12/2 + m2*v22/2 m1*v1 = m2*v2, so v2 = m1*v1/m2 so kx2 = m1*v12 +m2*(m1*v1/m2)2 =m1*v12 *(1+ m1/m2) v12 =kx2/[m1*(1 + m1/m2)] v22 = (m1/m2)2v12 =(m1/m2)2kx2/[m1*(1 +m1/m2)] m1*v12/2 = m1*g*d1, so v12 =2*g*d1 d1 = kx2/[2g*m1*(1 + m1/m2)] m2*v22/2 = m2*g*d2, so v22 =2g*d2 d2 = v22/(2g) =(m1/m2)2kx2/[2gm1*(1 + m1/m2)] xf = xi + d1 + d2 = xi = kx2[1 +(m1/m2)2]/[2gm1*(1 + m1/m2)] = 1.25 m
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