Homework Help: Question involving work and springs

1. Nov 3, 2012

mldavis086

1. The problem statement, all variables and given/known data

When a 4.00-kg object is hung vertically on a certain light
spring that obeys Hooke’s law, the spring stretches 2.50 cm.
If the 4.00-kg object is removed, (a) how far will the spring
stretch if a 1.50-kg block is hung on it? (b) How much
work must an external agent do to stretch the same spring
4.00 cm from its unstretched position?

2. Relevant equations

F = -k*x
W = F*x

3. The attempt at a solution

**I get a) correct so you can skip to b) if you want

a) First using the 4kg block I get a force of 39.2N. Then using 39.2 = -k*-0.025 and solving for k I get k= 1568 N/m. .
Then inputting k for the 1.5 kg block(F=14.7 N) I get 14.7=1568x. x=0.00938 m or x= 0.938 cm

b)This is what I tried to do: F=-1568*-0.04=62.8 N So the force needed is 62.8 N. So using W=F*x, I get W= 62.8*0.04 = 2.51 J

I can see in the back of the textbook the answer is 1.25 J, so I'm tempted to just divide my answer by 2 but then I won't understand why. Can anyone out there explain this to me please??

Last edited: Nov 3, 2012
2. Nov 3, 2012

voko

The force is not constant as the spring is stretched, so you need to integrate. Alternatively, if you know the formula for spring's potential energy, you could just use that.

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