A 0.6 kg collar slides freely along a curved rod. A massless linear spring is at
ID: 1828942 • Letter: A
Question
A 0.6 kg collar slides freely along a curved rod. A massless linear spring is attached
to the collar and the spring constant is k = 20 N/m. At point A the collar has a speed
of 2 m/s to the right and the distance between points A and O is 0.7 m. The
unstretched length of the spring is 0.1m and point A is 0.4 meters below point O.
a. Calculate the velocity of the collar just before it hits the wall at point B.
b. Calculate the force that the spring exerts on the collar at point B.
Explanation / Answer
(a)Total energy at point A=Kinetic Energy + Potential energy
Kinetic Energy= 0.5*0.6*2*2=1.2j
Potential Energy =0.5*k*x^2
P.E=0.5*20*0.6*0.6
P.E=3.6j
Total Energy at A=1.2+3.6=4.8j
Total energy at point B= Potential energy due o gravity + Potential energy stored in the spring+Kinetic energy
P.E due to gravity=0.6*9.8*0.4=2.352j
P.E stored in spring=0.5*20*0.3*0.3
P.E in spring =0.9j
By conservation of energy ,
total energies should be equal.
so, 4.8=2.352+0.9+K.E
K.E = 1.548j
0.5*0.6*v*v=1.548
v*v=1.548/0.3=5.16
velocity at top = sqrt(5.16)=2.271m/s
(b) At point B expansion in spring = 0.4-0.1=0.3
Force due to spring=k*x
Force = 20*0.3 = 6N
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