A mass is placed on an uncompressed spring and slowly lowered so that it comes t

Question

A mass is placed on an uncompressed spring and slowly lowered so that it comes to rest because it is in equilibrium (net force = 0). If the mass is 10.0 kg and the spring constant of the spring is 2500 N/m, how far is the spring compressed from its equilibrium position?

b) Now imagine that you raise the mass to a height of y = 6.0 cm above the top end of the uncompressed spring and drop it. It will fall and compress the spring, come to rest for an
instant and then bounce back up. First, determine the total mechanical energy of the
mass/spring system just before the mass is dropped. Measure the vertical distance y from
the top end of the uncompressed spring.

c) Next, write down an expression for the total mechanical energy of the mass/spring
system when the mass comes to rest momentarily on the compressed spring before
rebounding

Explanation / Answer

a) F=KX

X=F/K

X=Mg/K

X=(10)x9.8/2500

X=0.0392m

where X is the displacement from equilibrium position

b) just before mass is dropped on uncopmressed spring the spring-mass system has only potential energy of the mass which is

E=MgY

E=10x9.8x0.06

E= 5.88J

Vertical distance Y=6 CM

c) when the mass is dropped its total potential energy is converted into spring potential energy when it reached its equilibrium position which is 5.88J

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