Hooke's law with spring


by bearhug
Tags: hooke, spring
bearhug
bearhug is offline
#1
Sep28-06, 07:57 AM
P: 79
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. (a)If the object is replaced with one of mass 1.50 kg, how far will the spring stretch?
b) How much work must an external agent do (i.e. a force coming from the 'environment'), to stretch the same spring 4.00 cm from its unstretched position?

For a) I used the equation F= -kx to solve for k.
F=ma so ma=-kx since the object is vertical.
(4.00)(-9.8)= -k(.025m)
k= 1568 N/m

then I plugged it back in to solve for x
ma= -kx
(1.5)(-9.8)= -1568x
x= 0.00938 m

Is there something I'm doing wrong?
b) W= 1/2k(xi)^2 - 1/2k(xf)^2
W= 0-1/2(1568)(.04m)^2
= -1.25 J

The homework says this is wrong but I can't figure out what else to do?
Any help would be appreciated.
Phys.Org News Partner Science news on Phys.org
Lemurs match scent of a friend to sound of her voice
Repeated self-healing now possible in composite materials
'Heartbleed' fix may slow Web performance
Galileo
Galileo is offline
#2
Sep28-06, 10:14 AM
Sci Advisor
HW Helper
Galileo's Avatar
P: 2,004
Quote Quote by bearhug
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. (a)If the object is replaced with one of mass 1.50 kg, how far will the spring stretch?
b) How much work must an external agent do (i.e. a force coming from the 'environment'), to stretch the same spring 4.00 cm from its unstretched position?

For a) I used the equation F= -kx to solve for k.
F=ma so ma=-kx since the object is vertical.
(4.00)(-9.8)= -k(.025m)
k= 1568 N/m

then I plugged it back in to solve for x
ma= -kx
(1.5)(-9.8)= -1568x
x= 0.00938 m

Is there something I'm doing wrong?
This is correct. Another way to do this is to notice that Hooke's law means a linear relationship between F and x. So when you F is twice as big (in this case, if you make the mass twice as big), the spring is stretched twice as far. In this case, the mass is 1.5/4.0 times as heavy, so x is altered by the same factor:
[tex](1.50)/(4.00)*(2.5 cm)=0.938 cm[/tex]
However, your method gives the value for k, which you need in the next exercise.
b) W= 1/2k(xi)^2 - 1/2k(xf)^2
W= 0-1/2(1568)(.04m)^2
= -1.25 J

The homework says this is wrong but I can't figure out what else to do?
Any help would be appreciated.
The work done should be positive. You can see this by noting that the force you exert in the same direction as the displacement, or noting that by stretching a spring you increase the potential energy.
bearhug
bearhug is offline
#3
Sep28-06, 01:29 PM
P: 79
Thanks for responding. I'm using a computer program for this homework and it's telling me that both answers are wrong. This is after I saw what you wrote. So I'm just double checking to make sure that there is absolutely nothing wrong with this. It could be a glitch in the program.


Register to reply

Related Discussions
Ideal Spring and Real Spring Difference? (Hooke's Law) Introductory Physics Homework 10
[SOLVED] Spring Hooke's law Advanced Physics Homework 12
Hooke's law Introductory Physics Homework 1
difference between hooke's law and the work done in a spring? Introductory Physics Homework 3
Hooke's law with a spring Introductory Physics Homework 3