PHYSICS HELP PLEASE

Question

When a 3.80 kg object is hung vertically on a certain light spring that obeys Hooke's law, the spring stretches 2.80 cm.

 

(a) If the 3.80 kg object is removed, how far will the spring stretch if a 1.50 kg block is hung on it?

_________cm

 

(b) How much work must an external agent do to stretch the same spring 4.00 cm from its unstretched position?

_________J

Answer 1

Hi Porcia!

The spring, the spring stretches until the force the spring exerts up equals the weight of the mass.  In other words, until:

 

F_net-y = kx - mg = 0  (with kx up and mg down)

 

so  kx = mg, and k = mg/x = (3.8 kg)(9.8 m/s²)/0.028 m = 1330 N/m

 

What will be the new x value if a different mass, m = 1.5 kg, is hung from the spring?

 

In that case:

 

kx' = m'g --> (1330 N)(x') = (1.5 kg)(9.8 m/s²) --> x' = (1.5)(9.8)/1330 = 0.011 m or 1.1 cm (less then before, as expected with a smaller mass hung)

 

(b) The work done by an external agent to stretch the spring from unstretched (rest) length (x=0) to x = 4 cm = 0.04 m, is equal to the potential energy such an action stores in the spring.

 

U_s = (1/2)kx²

 

When the spring is unstretched, and x = 0, there is no energy stored.  When it is at x = 4 cm (= 0.04 m), it has stored:

 

U_s = (1/2)kx² = (1/2)(1330 N/m)(0.04 m)² = 1.064 J of potential energy is stored.  This is therefore the amoun of work an external agent must to to make that stretch and store that energy.

 

I hope this helps!  If you have any questions or problems with this, just let me know.