Physics Calculus Question

[Tacobell]

Hello, I have been working on these two simple calc questions and I am getting them wrong.

 

One is:

 

Suppose that 3 J of work is needed to stretch a spring from its natural length of 30 cm to a length of 42 cm.

 

(a) How much work W is needed to stretch the spring from 41 cm to 46 cm? (Round the answer to the nearest hundredth.)

 

For this one I am integrating (1/4)*k from 11 to 16 and getting 16.875 J

 

(b) How far beyond its natural length will a force of 50 N keep the spring stretched? (Round the answer to the nearest tenth.)

 

For this problem I am confused. I am just doing 50=(1/4)*x; this is incorrect too. I have a feeling that my problem on these two questions are units

 

There is this one last problem I am stuck on too, it seems very simple.

 

For this problem I just saw that my distance had a y value of 3f so that means 3*6*40m. What am I doing wrong here?

 

 

TacoBell

[Bignose]

Moved to Homework Help as this is the more appropriate section.

[timo]

What does k stand for, what it is value, how did you get this value? Note that usually the force is written as F = -k*displacement, not F=-(k/4)*displacement. But using k/4 consistently should still work fine in your case.

 

EDIT: Got the constant on the k wrong myself. should be fine, now.

Edited May 30, 2010 by timo

[swansont]

Work is [math]\int kx{dx}[/math], which gives [math] \frac{1}{2}kx^2[/math]

 

You're missing a factor of x, which will mess up your results beyond using 1/4 vs 1/2

[Tacobell]

What do you mean the "x" factor?

[the tree]

For this one I am integrating (1/4)*k from 11 to 16 and getting 16.875 J

 

Work is [imath]\int kx \,\mbox{d}x[/imath]...

You're missing a factor of x

 

What do you mean the "x" factor?

 

You should be integrating [imath]k x[/imath], not [imath]k[/imath]. Hence being out by a factor of [imath]x[/imath].