Calculating the Force of a Stretched Spring: A Scientific Approach

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Work of 5 Joules is done in stretching a spring from its natural length to 13 cm beyond its natural length. What is the force (in Newtons) that holds the spring stretched at the same distance (13 cm)?


here's what i done:

W = F * D
5 = F * 13
F = 5/13
F = kx
5/13 = k(13)
k = 5/39

\int_{0}^{13} 5/39 *x dx

i'm really confuse about this problem

 

[learningphysics]

Work of 5 Joules is done in stretching a spring from its natural length to 13 cm beyond its natural length. What is the force (in Newtons) that holds the spring stretched at the same distance (13 cm)?


here's what i done:

W = F * D
5 = F * 13
F = 5/13
F = kx
5/13 = k(13)
k = 5/39

\int_{0}^{13} 5/39 *x dx

i'm really confuse about this problem

Remember that the force depends on x. So using W=F*D does not work since the force is not constant. You need to use the integral.

Use:
W = \int_{0}^{13} k *x dx
to figure out k.

Then get the force.

 

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F = kx

from this the only thing i have is X, how would i found out k?

 

[learningphysics]

F = kx

from this the only thing i have is X, how would i found out k?

To get k, use the integral:
W = \int_{0}^{13} k *x dx

Solve the integral on the right side. What do you get?
Then substitude W=5 on the left side.

Now you should be able to get k.

 

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5=169/2*k
k = 10/169

would i have to setup another integral?

\int_{0}^{13} 10/169 *x dx

i get 5 if i solve that integral

 

[learningphysics]

5=169/2*k
k = 10/169

would i have to setup another integral?

\int_{0}^{13} 10/169 *x dx

i get 5 if i solve that integral

I'm sorry. You need to convert to meters.

So
W = \int_{0}^{0.13} k *x dx

I get k=591.7 N/m

Then F=kx=591.7* 0.13=76.92N

 

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thanks alot!