In the top image a 2.0 kg block is at rest in the equilibrium position on a fric
ID: 2002517 • Letter: I
Question
In the top image a 2.0 kg block is at rest in the equilibrium position on a frictionless surface attached to a spring with spring constant 345 N/m. The spring is compressed 10.6 cm from equilibrium and released. The lower image shows the mass in oscillation following its release.
Find the work done by the spring as the block moves from the compressed position to the equilibrium position.
What is the speed of the mass as it reaches the equilibrium point?
What is the kinetic energy of the block as it just reaches the equilibrium point after being released?
Explanation / Answer
a)
Work Done by the spring as the block moves form compressed position to equillibrium position is
W=PE=(1/2)KX2=(1/2)*345*0.1062
W=1.94 J
b)
Since the equillibrium point has no potential energy ,so all the potential energy will be equal to Kinetic energy,so
KE =1.94 J
(1/2)mV2 = 1.94
(1/2)*2*V2 =1.94
V=1.4 m/s
c)
Kinetic energy of the block at equilibrium point is
KE=1.94 J