Work, Power and EnergyEasy
Question
A block of mass m = 0.1 kg is connected to a spring of unknown spring constant k. It is compressed to a
distance x from rest. After approaching half the distance
from equilibrium position, it hits another block
and comes to rest momentarily, while the other block moves with a velocity 3ms-1. the total initial energy of the spring is
distance x from rest. After approaching half the distance
from equilibrium position, it hits another blockand comes to rest momentarily, while the other block moves with a velocity 3ms-1. the total initial energy of the spring is
Options
A.0.3 J
B.0.6 J
C.1.5 J
D.0.8 J
Solution
Apply momentum conservation
0.1u + m(0) = 0.1(0) + m(3)
0.1u2 = 
(3)2
kx2 = 
(0.1)32
Kx2 = 0.9
Kx2 = 0.6J
0.1u + m(0) = 0.1(0) + m(3)
0.1u2 = (3)2kx2 = (0.1)32Kx2 = 0.9Kx2 = 0.6JCreate a free account to view solution
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