# hooke's law help

• May 29th 2011, 03:40 AM
boromir
hooke's law help
A spring is attached to a block, mass m, on a frictionless ramp. Let ${\theta}$ be the angle the plane makes with the horizontal. Assume the spring has unstreched length t and constant k.

1) establish the forces acting on the block. I got normal reation, weight and the restoring force.

2) Find L,the length the spring streches to hold the block in equlibrium. I got $F-mgsin{\theta }=0$ so by hooke's law $L=mgsin{\theta }$ but the answer said I needed to add t. Can anyone explain why?

3)show the mechanincal energy of the block is given by $E=1/2(mx'^2)+mgsin{\theta x} +k/2(x^2)$. Where x=the displacement measured up the slopeand chosen such that x=0 when l=t

I know that E=PE +KE but don't know how to obtain the equation.

Thanks
• May 29th 2011, 08:26 AM
CaptainBlack
Quote:

Originally Posted by boromir
A spring is attached to a block, mass m, on a frictionless ramp. Let ${\theta}$ be the angle the plane makes with the horizontal. Assume the spring has unstreched length t and constant k.

1) establish the forces acting on the block. I got normal reation, weight and the restoring force.

2) Find L,the length the spring streches to hold the block in equlibrium. I got $F-mgsin{\theta }=0$ so by hooke's law $L=mgsin{\theta }$ but the answer said I needed to add t. Can anyone explain why?

What you have found is the extension (increased length) of the spring, the question asks for the length of the stretched spring.

CB
• May 29th 2011, 08:37 AM
CaptainBlack
Quote:

Originally Posted by boromir

3)show the mechanincal energy of the block is given by $E=1/2(mx'^2)+mgsin{\theta x} +k/2(x^2)$. Where x=the displacement measured up the slopeand chosen such that x=0 when l=t

I know that E=PE +KE but don't know how to obtain the equation.

Thanks

The three terms are the KE of the block, the change in gravitational potential energy of the block, and the (potential) energy stored in the stretched spring.

CB
• May 29th 2011, 11:41 AM
boromir
I can derive the stored PE.
Can you help me derive the others please?
• May 29th 2011, 10:04 PM
CaptainBlack
Quote:

Originally Posted by boromir
I can derive the stored PE.
Can you help me derive the others please?

The first is the kinetic energy (1/2)mv^2, the second is the loss of potential energy mgh=mg sin(theta)x, where x is negative, the third is the work done in extending the spring which is the integral of kx from 0 to x.

CB
• May 30th 2011, 07:31 AM
boromir
Thanks though I am having a bit of trouble with the signs. If we take up the slope as positive, then we have stored energy= $-{\int}-kx=1/2(kx^2)$ , loss of $GPE=(-mg)(-x)sin{\theta }$and KE=1/2(mv^2). Is that correct?

The next question is what is the total mechanical energy when the spring is pushed up a distance y from the unstreched length. My answer is $1/2(ky^2)-mgysin{\theta}+1/2(mv^2)$, again taking up the slope as the positive direction.

Then it asks when the cart is released from rest at position y, what is its speed when the spring has returned to its unstretched length t? Initally $E=-mgysin{\theta}+1/2(ky^2)$ and finally $E=1/2mv^2$. Equating the two yields v= ${\sqrt}(ky^2-2mgysin{\theta})/m$.

Am I correct. Again thanks
• May 30th 2011, 07:54 AM
CaptainBlack
Quote:

Originally Posted by boromir
Thanks though I am having a bit of trouble with the signs. If we take up the slope as positive, then we have stored energy= $-{\int}-kx=1/2(kx^2)$ , loss of $GPE=(-mg)(-x)sin{\theta }$and KE=1/2(m(-v)^2)=1/2(mv^2). Is that correct?

The next question is what is the total mechanical energy when the spring is pushed up a distance x1 from the unstreched length. My answer is [tex]1/2(kx1^2)

The sign of the change in potential energy should be negative, the block has moved down the incline so $x$ is negative hence

$mgx \sin(\theta)$

is negative as required.

In the other terms $x$ appears squared and so there is no sign problem.
• May 30th 2011, 08:35 AM
CaptainBlack
Quote:

Originally Posted by boromir

The next question is what is the total mechanical energy when the spring is pushed up a distance y from the unstreched length. My answer is $1/2(ky^2)-mgysin{\theta}+1/2(mv^2)$, again taking up the slope as the positive direction.

The change in potential energy is:

$mgy \sin(\theta)$

now the sign of $y$ is positive but before $x$ was negative.

CB
• May 30th 2011, 08:40 AM
CaptainBlack
Quote:

Originally Posted by boromir
Then it asks when the cart is released from rest at position y, what is its speed when the spring has returned to its unstretched length t? Initally $E=-mgysin{\theta}+1/2(ky^2)$ and finally $E=1/2mv^2$. Equating the two yields v= ${\sqrt}(ky^2-2mgysin{\theta})/m$.

Am I correct. Again thanks

Initially the cart is at rest so there is no KE. The potential energy of the system is:

$E_0=mgy \sin(\theta) + \frac{ky^2}{2}$

When the spring has returned to it's natural length there is only KE and it is equal to:

$KE=E_0$

so:

$(1/2)mv^2=mgy \sin(\theta) + \frac{ky^2}{2}$

which we solve for $v$

CB
• May 30th 2011, 09:29 AM
boromir
with regards to the sign of mgysin(theta) the sign should be negative because mg is negative whearas y is positive so you would have (-mg)ysin(theta)= -mgysin(theta).
• May 30th 2011, 11:36 AM
CaptainBlack
Quote:

Originally Posted by boromir
with regards to the sign of mgysin(theta) the sign should be negative because mg is negative whearas y is positive so you would have (-mg)ysin(theta)= -mgysin(theta).

Slightly different convention, but yes if g is signed, because this PE has to be positive for a displacement up the incline.

CB
• May 30th 2011, 12:21 PM
boromir
Can you explain a bit clearer please?
• May 30th 2011, 12:30 PM
CaptainBlack
Quote:

Originally Posted by boromir
Can you explain a bit clearer please?

$g$ is a constant that you do not want to depend on your reference frame. The minute you say $g=-9.81 \text{ m/s}^2$ you have partially specified a reference frame. In fact we do not have a reference fame of this sort here we just have a single coordinate for displacement which is positive up the incline.

We need the change in PE for a displacement $y$ up the incline to be positive so if we have an unsigned g the change in PE is $mgy \sin(\theta)$. If we use a negative signed $g$ we will have the change in PE as $-mgy \sin(\theta)$, but now we have to worry about the orientation of the coordinates in a 2 or 3D space containing our inclined plane.

In fact it is usual to define g unsigned (see the Wikipedia page on standard gravity for an example)

CB
• May 31st 2011, 04:28 AM
boromir
Thanks a lot for your help