Total Elastic Potential Energy

[Zurtex]

This is not my problem, it's a friends but I have no idea how to tackle it:

A block B of mass 5 kg is fastened to one end of each of two springs. The other ends of the springs are attached to fixed points A and C, 4 metres apart on a smooth horizontal surface, as shown in the diagram.

Spring AB has natural length 2 metres and modulus of elasticity 30 N, while BC has natural length 1 metre and modulus 40 N (you may assume that the springs meet at the centre of B).

Find the total elastic potential energy of the system in this position.

He knows it is something to do with:

$$\frac{\lambda_{AB} X_{AB}}{l_{AB}}=\frac{\lambda_{BC} X_{BC}}{l_{BC}}$$

Can you help please in pointing him in the right direction (I'll give him the link to this thread)

 

[daster]

Elastic potential energy (EPE) is given by the equation:

$$EPE = \frac{\lambda}{2l}x^{2}$$

where $\lambda$ is the modulus of elasticity, $l$ is the natural length and $x$ is the extension of the spring.

For this question, consider the springs AB and BC seperately. You have the modulus and natural length of both springs, so all you need to find is the extension. Seeing as the system is in equilibrium, then the tension in spring AB is the equal in magnitude to the tension in spring BC. Tension (T) is given by the equation:

$$T=\frac{\lambda}{l}x$$

where the constants represent the same values as in EPE's equation.

Remember to take into consideration that the length of both springs is 4. You should now have a bunch of simultaneous equations.

 

[Bucky]

Hi, could you explain what you mean by the extension of the spring (value for x)?

Also for the second formula, how would you incorporate the fact that the total length is 4m? would you do an equation using the sum of the values for AB and BC?

 

[daster]

Consider a spring, XY, of natural length 1m. If you pull that spring until it's length is, say, 1.5m, then its extension is 0.5m.

The total length of AB is 2+x and of BC is 1+y (where x & y are the extensions of AB & BC and 2 & 1 are their natural lengths, respectively). Since the total length of AC is 4, then: AB+BC=4.

 

[Bucky]

Ok here's what i have so far...it look ok?

$$EPE_A_B = \frac{\lambda}{2l} x^2 = \frac{3}{4} 0.5^2 = 0.1875$$

or am i way off? i thought since the total length is 4m, and the sum of the spring lenths is 3m, and they are in equilibrium, that the extension of the spring would be the total length, minus the spring lengths, divided by the two (ie 0.5)

 

[daster]

Hmm, nope, you're way off. :tongue2:

First we have:
2+x+1+y=4 => x+y=1

Second, consider the tension the the two springs:
(Remember since the system is in equilibrium, the tension in spring AB is equal to the tension in spring BC.)

$$T_{AB} = T_{BC}$$

$$\frac{\lambda_{AB}}{l_{AB}}x = \frac{\lambda_{BC}}{l_{BC}}y$$

$$\frac{30}{2}x=\frac{40}{1}y$$

And those are your two simultaneous equations. Solve them to get x and y.

You have to realize that the springs have different extensions (lengths and moduli, too), so they must be considered seperately. Once you have the values of x and y, obtain EPE for spring AB (using the value of x) and EPE for spring BC (using the value of y). For the total EPE, add these two up.

 

[Bucky]

well here we go...
x + y = 1
15x - 40y = 0

15x + 15y = 15
15x - 40y = 0

(2) - (1)

-65y = -15
65y = 15
y = 65/15
y = 3/13

x + y = 1
x + 3/13 = 13/13
x = 13/13 - 3/13
x = 10/13

$$EPE_{AB} = \frac{\lambda}{2l} x^2 = \frac{3}{4} X \frac{3}{13}^2 = \frac{3}{4} X \frac{9}{169} = \frac{27}{676} = 0.0399 \\$$


$$EPE_{BC} = \frac{\lambda}{2l} x^2 = \frac{4}{2} X \frac{10}{13}^2 = 2 X \frac{100}{169} = \frac{200}{169} = 1.1834 \\$$


$$EPE_{TOTAL} = 0.0399 + 1.1834 = 1.2233$$

i get the feeling I am way off...but i allways get that feeling...

 

[daster]

Your method is correct, but your calculations are a bit off.

I got y=3/11 and x=8/11.
EPE_ab = (30/4) x^2 = 3.97
EPE_bc = (40/2) y^2 = 1.49
EPE_total = 3.97 + 1.49 = 5.46 J