What is the Work Required to Stretch a Spring?

[blayman5]

Homework Statement

A spring with spring constant of 29 N/m is
stretched 0.21mfromits equilibrium position.
How much work must be done to stretch it
an additional 0.14 m? Answer in units of J.

Homework Equations

W=-kx(x)/2

The Attempt at a Solution

I solved it and got 0.4263 and it was wrong. does it need to be negative?

 

[LowlyPion]

Homework Statement

A spring with spring constant of 29 N/m is
stretched 0.21mfromits equilibrium position.
How much work must be done to stretch it
an additional 0.14 m? Answer in units of J.

Homework Equations

W=-kx(x)/2

The Attempt at a Solution

I solved it and got 0.4263 and it was wrong. does it need to be negative?

How did you arrive at that number?

 

[blayman5]

i did ((29)(0.21)(0.14))/2 =0.4263. does it need to be negative?

 

[LowlyPion]

i did ((29)(0.21)(0.14))/2 =0.4263. does it need to be negative?

That's not quite right.

Additional work = Difference in work = 29*(.21+14)²/2 - 29*(.21)²/2

 

[blayman5]

Ok so that means it would equal
29*(.21+14)2/2 - 29*(.21)2/2 =1.1368

Thanks

 

[LowlyPion]

Ok so that means it would equal
29*(.21+14)2/2 - 29*(.21)2/2 =1.1368

Thanks

That's what I got anyway. What you need to remember is that work is the area under the force and distance curve over the range of the curve.

Cheers

 

[blayman5]

i will remember thanks

 

[dleccord]

That's not quite right.

Additional work = Difference in work = 29*(.21+14)²/2 - 29*(.21)²/2

wait...

wait wait... why is it 14 instead of .14?

 

[LowlyPion]

wait...

wait wait... why is it 14 instead of .14?

He got the right answer and the 14 was apparently a typo. Calculate it out and you will see the inner sum needs to sum to .35.