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Problem with box and string

by GingerKhan
Tags: string
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GingerKhan
#1
Nov13-11, 11:58 AM
P: 14
1. The problem statement, all variables and given/known data

A spring with a force constant of 225 N/m is resting on a friction-less surface and mounted against a wall horizontally. A 1.5 kg box is pushed against the spring and compresses it 12 cm (0.12m) from equilibrium. When released the spring pushes the box across the surface.


2. Relevant equations

F = kx

W = FΔd

Ee = 1/2 k x^2

Ek = 1/2 m v^2

3. The attempt at a solution

a) How much force needs to be applied to the spring to compress it to 12 cm (0.12m)?

F = 225 x 0.12 = 27 N


b) How much work is done to compress the spring to 12 cm?

W = 27 x 0.12 = 3.24 J


c) How much elastic energy is stored in the spring when compressed?

Ee = 1/2 x 225 x 0.12^2 = 1.62 J

Question: Is it normal that Ee is less than the work energy applied to the system in b)?


d) What maximum speed will the box attain once released?

1.62 = 1/2 x 1.5 x v^2

1.62/0.75 = v^2

v = 1.47 m/s

** not sure about this because I might be using a wrong Ek obtained in c) **


Thanks in advance.
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grzz
#2
Nov13-11, 12:08 PM
P: 950
[QUOTE=GingerKhan;3613702]
... How much work is done to compress the spring to 12 cm?

W = 27 x 0.12 = 3.24 J


c) How much elastic energy is stored in the spring when compressed?

Ee = 1/2 x 225 x 0.12^2 = 1.62 J

Question: Is it normal that Ee is less than the work energy applied to the system in b)? ...

Note that the force doing the work is varying and so an AVERAGE value of the force must be taken.
Filip Larsen
#3
Nov13-11, 12:09 PM
PF Gold
Filip Larsen's Avatar
P: 958
Quote Quote by GingerKhan View Post
b) How much work is done to compress the spring to 12 cm?

W = 27 x 0.12 = 3.24 J
This is not correct. You can only use "work equals force times distance" on the whole distance if both are constant. If they are not you must integrate the work up from the instantaneous force as a function of distance, e.g. W = ∫F(s)*ds, where s is distance.

In the context of this problem you may also solve b) by using conservation of energy. The original kinetic energy from the block must all be stored in the spring since there is no friction and nothing else moves, hence the block does the amount of work on the spring equal to its kinetic energy, which again is equal to the potential energy right after compression.

cepheid
#4
Nov13-11, 12:11 PM
Emeritus
Sci Advisor
PF Gold
cepheid's Avatar
P: 5,196
Problem with box and string

Welcome to PF GingerKhan!

Quote Quote by GingerKhan View Post
1. The problem statement, all variables and given/known data

A spring with a force constant of 225 N/m is resting on a friction-less surface and mounted against a wall horizontally. A 1.5 kg box is pushed against the spring and compresses it 12 cm (0.12m) from equilibrium. When released the spring pushes the box across the surface.


2. Relevant equations

F = kx

W = FΔd

Ee = 1/2 k x^2

Ek = 1/2 m v^2

3. The attempt at a solution

a) How much force needs to be applied to the spring to compress it to 12 cm (0.12m)?

F = 225 x 0.12 = 27 N


b) How much work is done to compress the spring to 12 cm?

W = 27 x 0.12 = 3.24 J


c) How much elastic energy is stored in the spring when compressed?

Ee = 1/2 x 225 x 0.12^2 = 1.62 J

Question: Is it normal that Ee is less than the work energy applied to the system in b)?
Nope. These two numbers should be the same. The mistake is in assuming that W = FΔx. This is only true if the force is constant. The force is not constant here, but rather it varies with x. The more general formula is W = ∫F(x)dx where F(x) is the (position-varying) force function. But if you don't know integral calculus, don't worry. In this case, you've already (implicitly) been given the expression for the work done. Think about this: what is the work done by the spring force as the spring is compressed, (hint: how does this relate to the elastic potential energy stored)? Now, how does the work done by the applied force have to compare to the work done by the spring force?
grzz
#5
Nov13-11, 12:13 PM
P: 950
Since the force F is directly proportional to the compression the average value of the force will be (final value of F)/2.
GingerKhan
#6
Nov13-11, 01:11 PM
P: 14
Oh, I forgot about the spring becoming harder to compress as it is being compressed.

So with a graph that illustrates the scenario, I would need to find the area of the right triangle and divide the whole thing by two, giving me 1.62 J

I could also use Ee = 1/2 k x^2 to calculate the work done because W = ΔE.

If c) is correct then d) is correct as well, right?

Thanks guys.


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