# Believe it or not, I am able to do a lot of this stuff without the board

It's just that it helps to understand what I'm doing.

4. [HRW6 7.PN.04.] The block in Fig. 7-10a lies on a horizontal frictionless surface and is attached to the free end of the spring, with a spring constant of 65 N/m. Initially, the spring is at its relaxed length and the block is stationary at position x = 0. Then an applied force with a constant magnitude of 3.3 N pulls the block in the positive direction of the x axis, stretching the spring until the block stops.
Assume that the stopping point is reached.

Okie doke..

When the block stops, its acceleration is zero. When the acceleration is zero, the net force must be zero. So Fapplied must equal the spring force. The applied force is 3.3, so then I can use Hooke's law

Fspring = -Kx
-3.3 = -65x
3.3/65 = .0507 m

What did I do wrong?

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Doc Al
Mentor
Note that the applied force is constant. This is not the same thing as gently stretching the spring with $F_{applied} = kx$ until some specified force is reached. Hint: Consider the work done by the applied force.

Well ok, how can I find the work done if i dont know the distance it's been pushed?

The work done by a spring is $W = \frac{1}{2}kx^2$
The work done by the E field is $W = qE\Delta x$

We're not doing the Electric Fields yet, this should just be simple mechanics...

So should I do Fapplied(x)=1/2kx^2?

Wow, stupid me. I posted in the wrong thread. The work done by both will be equal, I think thats what your post says.

And I thought you were my friend!

if it's lying horizontal and the spring is initially at rest the stretched to x+d. Is the spring let go and allowed to oscillate back and forth, is the block attached to it. If it isn't then the spring won't push it any since it can only oscillate to x(+,-)d.

Its like you pulling at an object on a spring. The block is attached to the spring and experiences an external force of 3.3N

edit: the force isnt external, because it only acts on the block. The force of 3.3N acts on the block, and in turn stretches the string.

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learningphysics
Homework Helper
Doc Al pointed out the error. I'd just add that your mistake was that you took acceleration=0, when you actually need velocity=0 (that's the stopping point). Work done goes completely into potential energy in the spring, and 0 kinetic energy.

3.3*x = (1/2)kx^2

Can't you take Fapplied=kx solve for x since you know k and Fapplied. take work W=1/2kx^2 integrate from 0 to whatever distance it was pulled.

qqchico said:
Can't you take Fapplied=kx solve for x since you know k and Fapplied.
That would be the same mistake the OP did.

take work W=1/2kx^2 integrate from 0 to whatever distance it was pulled.
Integrating force from 0 to x would give the correct amount of work done, but since force is constant, then the integral would just simplify to F_applied * x

Integrating work would make no sense (in this case).

No it wouldn't 1/2*k are constant pull that out your left with x^2 integrate that you'll get (1/3)x^3*(1/2)k. would it be useful in this case don't know.

whozum said:
Integrating work would make no sense (in this case).

No I said "take work W=1/2kx^2 integrate from 0 to whatever distance it was pulled." No biggie though were probably more worked up about the problem than the original poster. Do you know any quantum. I need help with that.

qqchico said:
No I said "take work W=1/2kx^2 integrate from 0 to whatever distance it was pulled." No biggie though were probably more worked up about the problem than the original poster. Do you know any quantum. I need help with that.
You're still missing my point but nevermind. Start a new thread for your QM issues.

We'll then how do you solve it? Do you set work =fapplied and solve for x.

learningphysics
Homework Helper
qqchico said:
We'll then how do you solve it? Do you set work =fapplied and solve for x.
Fapplied * x = (1/2) kx^2

3.3x = (1/2) kx^2

Solve for x.

Theres no net change in velocity, to the net work done is zero. The work done by the spring is .5kx^2, and the work done by the constant applied force is Fx.

$$2Fx = kx^2$$

$$2F = kx$$

$$x = \frac{2F}{k}$$

$$x = \frac{2(3.3)}{65} = 0.102m$$

I see ok :rofl:

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OlderDan
Homework Helper
whozum said:
Theres no net change in velocity, to the net work done is zero. The work done by the spring is .5kx^2, and the work done by the constant applied force is Fx.

$$2Fx = kx^2$$

$$2F = kx$$

$$x = \frac{2F}{k}$$

$$x = \frac{2(3.3)}{65} = 0.102m$$
and just to tie it together, this result is of course twice the distance calculated by the OP who was actually calculating the displacement from the starting point to equilibrium. Under the influence of the combined forces of the spring and the constant applied force, the mass moves from rest, through equilibrium, to a maximum displacement on the other side of equilibrium, then returns to its original position and continues to oscillate about equilibrium at the midpoint of its motion.

Relative to the equilibrium position, the work done by the combined forces is $$\frac{1}{2}ky^2$$ where y is the displacement from equilibrium. So you have the usual energy conservation of an undamped oscillator.