A #3.0# #kg# bag of potatoes is sitting on top of a spring with a spring constant #k=600# #Nm^-1#. (a) What is the compression of the spring at the equilibrium point?

The sack is lifted by #0.5# #m# above the spring and dropped on it. (b) What is the maximum compression of the spring?

1 Answer
Nov 2, 2017

a) The compression of the spring at equilibrium is #0.049# #m#

b) The maximum compression of the spring is #0.22# #m#

Explanation:

a) The weight force of the bag of potatoes is given by:

#F=mg=3xx9.8=29.4# #N#

The Hooke's law expression for the spring is #F=kx#. Rearrange to make #x# the subject:

#x=F/k=29.4/600=0.049# #m = 4.9# #cm#

b) We need to calculate the kinetic energy of the falling bag just before it hits the spring, then convert that kinetic energy into spring potential energy:

#E_k=1/2mv^2#

But we don't yet know the velocity of the bag:

#v^2=u^2+2as=0^2+2xx9.8xx0.5=9.8#

Therefore #v=sqrt(9.8)=3.1# #ms^-1#

Then:

#E_k=1/2mv^2=1/2xx3xx3.1^2=14.4# #J#

This converts to spring potential energy:

#E_(sp)=1/2kx^2#

Rearrange to make #x# the subject:

#x=sqrt((2E_(sp))/k)=sqrt((2xx14.4)/600)=0.22# #=22 cm#