Calculating Spring Constant of a Hanging Spring

[teng125]

A spring with length l = 8 cm hangs in
vertical direction somewhere close the Earth's surface. When a mass of m = 10 kg is attached
to the spring its rest-position is at y0 = 11 cm. (a) Determine the spring constant k.

i try to use F=-kx = mg where x=11cm - 8cm
therefore,i would like to know whether my steps are correct or i used the wrong formula

somebody pls help

 

[teng125]

part b) The
mass is pulled out of its rest-position by a distance Δy = 5 cm. What is the acceleration acting
on the mass in the moment of its release?

i got the answer of a=16.35m/s^2

am i right??

 

[sdekivit]

A spring with length l = 8 cm hangs in
vertical direction somewhere close the Earth's surface. When a mass of m = 10 kg is attached
to the spring its rest-position is at y0 = 11 cm. (a) Determine the spring constant k.

i try to use F=-kx = mg where x=11cm - 8cm
therefore,i would like to know whether my steps are correct or i used the wrong formula

somebody pls help

You're in the right direction :) Remember that at the spring its rest position F_{gr.} = F_{spring} = C \cdot u where C = spring constant.

So you need to solve: m \cdot g = C \cdot u with u = distance of strecthing of the spring (3 cm)

 

[sdekivit]

foir b i would say that you use the second law of Newton: F_{res} = m \cdot a

 

[Doc Al]

i try to use F=-kx = mg where x=11cm - 8cm
therefore,i would like to know whether my steps are correct or i used the wrong formula

You are correct. (Except for that minus sign. Set the magnitude of the spring force equal to the magnitude of the weight.)

 

[Doc Al]

part b) ...
i got the answer of a=16.35m/s^2

am i right??

Looks OK to me. (But next time don't just give your answer; show how you got your answer. )

 

[sdekivit]

agrees with Doc Al :)