Mechanical Physics

Question

You have a light spring which obeys Hooke's law. This spring stretches 2.76 cm vertically when a 2.10 kg object is suspended from it. Determine the following.

(a)

the force constant of the spring (in N/m)

 ___  N/m

(b)

the distance (in cm) the spring stretches if you replace the 2.10 kg object with a 1.05 kg object

 __cm

(c)

the amount of work (in J) an external agent must do to stretch the spring 6.90 cm from its unstretched position

 __J

Answer 1

Hey John,

a) You should be familiar with Hooke's law - that the amount that a spring stretches is proportional to the applied force. The "force constant of the spring", is just the force on the spring divided by the amount that the spring stretched. Here, you'll need those numbers in Newtons and meters (respectively). We can calculate the force of gravity due to the mass with F = mg. And for the stretch of the spring, just convert it to meters. Divide and you are done with part a.

b) You can save a little math on this one by realizing that the new mass is half the original mass. Since this is a spring that obeys Hooke's law, the spring will stretch exactly half as much when there is half as much mass hanging from it.

c) With work/energy, things can get confusing because there are A LOT of energy formulas. So you have to ask, "where is the work going?". In this case, the work is all going into stretching a spring, which gives it elastic potential energy. So you can just use your formula for potential energy in a spring:

1. PE_spring = 1/2 k*x²
2. Just remember that, since N/m, you must first put your x value in meters instead of cm. Your answer will be in units of N*m, which is equivalent to Joules.