# Pogo Stick and Spring Constants

1. Oct 8, 2006

### techninja

A child's pogo stick (see figure) stores energy in a spring of spring constant k = 40000 N/m. At position A, x1 = -0.150 m, the spring compression is a maximum and the child is momentarily at rest. At position B, the spring is relaxed and the child is moving upward. At position C, the child is again momentarily at rest at the top of the jump. Assuming that the combined mass of child and pogo stick is 26.0 kg, answer the following:

(a) Calculate the total energy of the system if both potential energies are zero at x = 0

I did this through:
$$ME_i = U_s + U_g = \frac{1}{2}kx^2+mgh = 411.78 J$$

(b) Determine x2

I did this through:
$$ME_i = ME_f = U_g ME_i = mgh$$
$$x_2 = \frac{ME_i}{mg} = 1.62 m$$

(c) Calculate the speed of the child at x = 0

I did this through:
$$ME_i = \frac{1}{2}mv^2 v = \sqrt{\frac{2ME_i}{m}} = 5.63 m/s$$

(d) Determine the value of x for which the kinetic energy of the system is a maximum.
Hint: The kinetic energy (and velocity) is a maximum when the acceleration is zero at that point.

This one, I tried very hard... but I can't do it.

I get:
$$v = \sqrt{\frac{2(1378-mgh)}{m}} a = \frac{1}{m\sqrt{\frac{2h}{m}}}$$

Tex hates me. It should look something like v = sqrt([2(1378-mgh)]/m), so a = 1/(m*sqrt(2h/m))

Which doesn't seem to work out nicely.

(e) Obtain the child's maximum upward speed

I think I could do this with information from D, but ... I'm not sure.

Any help is muchly appreciated!

Last edited: Oct 8, 2006
2. Oct 8, 2006

### Staff: Mentor

This is easier than you think. First realize that as soon as the system passes x = 0, the speed can only decrease. So the max speed must occur before that point. Use your expression for total ME as a function of position and solve for the value of x that maximizes KE. (For KE to be max, PE must be min.)

For an even easier solution, use your own idea that the acceleration must be zero: For the acceleration to be zero, the net force must be zero. Where is that point?

3. Oct 8, 2006

### techninja

I'm not sure what you mean. If I try to find position, I end up with:

$$KE = ME_i - mgh$$

Where ME_i is a constant, as are m and g. Thus, I get a linear equation.

For the latter comment, I'm not sure how the free-body diagram should then be drawn as. There's force of gravity going downwards, but isn't kinetic energy upwards in units of joules? How can I change that?

4. Oct 8, 2006

### Staff: Mentor

Don't forget spring PE, which is also a function of position.

(1) KE is not a force. And energy is a scalar, not a vector--it has no direction. (You probably mean that the velocity is upwards, but that's also irrelevant.)
(2) There are two forces acting on the system. Gravity is one. What's the other?

5. Oct 9, 2006

### techninja

Thank you verily much! It suddenly makes sense! (I was wondering why the spring constant was perfectly related to part d)

I still don't understand the second approach, however. Is it possible to find the other force? (Which I assume is just... force imparted upwards?)

6. Oct 9, 2006

### Staff: Mentor

Of course it is. What is exerting the upward force on the system? (Hint: It's the same thing you missed with the first approach!)

Note: The two approaches are just different ways of doing the same thing.