Potential Energy/Work done question AS Level Physics - Mechanics

[novamatt]

Homework Statement

The question posed - Calculate the work done in a steel wire (Esteel = 200GPa) 2.0m long and 0.010cm² in cross-sectional area extends by 2.5mm when loaded.

so

Length (L) = 2m
Area (A) = 1(x10 to the -6)m
Extension (x) = 0.0025m
Ep? = 2(x10 to the 11) Nm?

Homework Equations

I know work done = Force x distance and EP = 1/2F$\chi$

The Attempt at a Solution

my attemp at solving this as follows:

1/2 2(x10 to the eleven)Nm x 0.0025m
= 1(x10 to the eleven)Nm x 0.0025m
= 2.5(x10 to the seven) Nm²

this just does not seem right to me... I haven't used half of the information give in the problem and I'm pretty sure potential energy is not neccissarly the same as work done. Also if the was the case wouldn't the spring constant reach it's ultimate tensile stress or fracture point.

Please help I was away when we covered this in college and the notes provided do not seem to help at all.

 

[PhanthomJay]

Homework Statement

The question posed - Calculate the work done in a steel wire (Esteel = 200GPa) 2.0m long and 0.010cm² in cross-sectional area extends by 2.5mm when loaded.

so

Length (L) = 2m
Area (A) = 1(x10 to the -6)m
Extension (x) = 0.0025m

yes

Ep? = 2(x10 to the 11) Nm?

The modulus of elasticity (Esteel) is an inherent property of the material (in this case, steel) and is a measure of its rigidity, with units of N/m^2

Homework Equations

I know work done = Force x distance

only if the force is constant and the displacement is in the direction of the force

and EP = 1/2F$\chi$

yes, OK

The Attempt at a Solution

my attemp at solving this as follows:

1/2 2(x10 to the eleven)Nm x 0.0025m
= 1(x10 to the eleven)Nm x 0.0025m
= 2.5(x10 to the seven) Nm²

this just does not seem right to me... I haven't used half of the information give in the problem

And what you have used is not right, ..you have incorrectly equated F with E

and I'm pretty sure potential energy is not neccissarly the same as work done.

It's the negative of the change in PE that is the same as work done by a conservative force (like a spring, which is the case here)

Also if the was the case wouldn't the spring constant reach it's ultimate tensile stress or fracture point.

Not if you use the proper values and the steel stays within its elastic limit

Please help I was away when we covered this in college and the notes provided do not seem to help at all.

What you have mostly missed is the equation for the extension of the steel under an applied axial load. If $\chi$ is the extension, then it can be shown that


$\chi$ = FL/AE
or
F= (AE/L)$\chi$

Since the steel member obeys Hooke's law, F=k$\chi$, then

k=AE/L

And you can solve for the work done using the same formula you use when assuming the steel rod is a spring (or use your equation W = F/2($\chi$)).

 

[novamatt]

thank you for those equations they are not in my notes anywhere... and thank you for taking the time to help me with my problem

so...

F =
1(x10 to the -6)m² x 2(x10 to the 11)Nm / 2m = 100000Nm
100000Nm x 0.0025m
= 250 Nm²

Work Done =
250Nm² x 0.0025m
= 0.625 J

still this somehow does not seem right? ie. the units do not seem to match up for me or am I missing something?

you mentioned that work done = force x distance only if force is a constant what would happen if this was not the case?

 

[PhanthomJay]

thank you for those equations they are not in my notes anywhere... and thank you for taking the time to help me with my problem

You're welcome.

so...

F =
1(x10 to the -6)m² x 2(x10 to the 11)Nm / 2m = 100000Nm
100000Nm x 0.0025m
= 250 Nm²

Watch your units...E has units of N/m^2, so the force is 250 N.

Work Done =
250Nm² x 0.0025m
= 0.625 J

No, the force is not constant...it varies from 0 as the wire just starts to stretch, to 250 N when it is fully stretched...the work done is the average force times the extension...W = (250/2)(.0025) = .3125 J

you mentioned that work done = force x distance only if force is a constant what would happen if this was not the case?

As I noted, this is not the case...when the force is not constant, and the force is in the direction of the displacement, then
$W = \int F.dx$
I don't know if you have taken calculus, but since F =kx, per Hookes Law, then $W =\int kx.dx = 1/2kx^2$. Or if you don't know calculus, you can use, in this case, since the force varies linearly with the distance, the other formula, $W = (F/2)(x) = (kx/2)(x) = 1/2kx^2$, the same result.

Note that the work done by the force is the same value as the potential energy change of the steel wire (a spring), but I don't know if you have studied energy methods yet.

 

[novamatt]

thanks I see now that you had already explained the force was not a constant and can now see where the equation in my notes 1/2kx² comes into play. =) I shall return to you for pointers in the future you've been a great help