If the length of a #65 cm# spring increases to #94 cm# when a #5 kg# weight is hanging from it, what is the spring's constant?

1 Answer
Aug 18, 2016

Answer:

Draw a free body diagram first

Explanation:

Let's sort this through.

A #5kg# comes to equilibrium with the spring and since the box is not accelerating in either direction, the net force is zero. We would set the weight of the box equal to the force on the spring aka the restoring force

#Ho##oke's# #law# states:

#F = -kx#

where #k# is the spring constant in #N/m#
and #x# is the change in displacement of the spring from the equilibrium position in #m#

*We can ignore the (-) sign in this case because that just indicates the force is a restoring force.

Setting up the forces to equal each other, we get:

#kx = m*g#
#k = (m*g)/x#
#k = ((5 kg) * (9.8 m/s^2))/(0.94m-0.65m)#
#k = 168.97 N/m#