- #1

Argonaut

- 45

- 24

- Homework Statement
- An experimental apparatus with mass ##m## is placed on a vertical spring of negligible mass and pushed down until the spring is compressed a distance ##x##. The apparatus is then released and reaches its maximum height at a distance ##h## above the point where it is released. The apparatus is not attached to the spring, and at its maximum height it is no longer in contact with the spring. The maximum magnitude of acceleration the apparatus can have without being damaged is ##a##, where ##a > g##. (a) What should the force constant of the spring be? (b) What distance ##x## must the spring be compressed initially?

- Relevant Equations
- $$F=ma$$

$$U_{\text{grav}}=mgh$$

$$U_{\text{el}}=\frac{1}{2}kx^2$$

Here is

a) The apparatus may only experience acceleration ##a > g## while in contact with the spring. Since the spring exerts the greatest force when it is the most compressed, the apparatus will undergo the greatest acceleration at that point. So Newton's second law gives

$$\sum F = ma$$

$$kx-mg = ma$$

Therefore, the force constant of the spring should be $$k = \frac{m(a+g)}{x}$$.

b) There are only conservative forces in the system, so energy is conserved. Let point 1 (with ##y=0##) be the point where the apparatus is released and let point 2 be the point where it reaches height ##h##. Then

$$U_1=U_2$$

$$\frac{1}{2}kx^2 = mgh$$

Expressing ##x##

$$x=\sqrt{\frac{2mgh}{k}}$$

However, the

a)

$$k = \frac{m(g+a)^2}{2gh} $$

b)

$$x = \frac{2gh}{g+a} $$

I could 'reverse-engineer' both solutions. However, I don't understand how I should have known to express ##k## in terms of ##m##, ##a##, ##g## and ##h##, and not ##x##. Is it because of part b? Because essentially, both ##k## and ##x## are target variables and only the rest are known?

**my attempt**at the solution:a) The apparatus may only experience acceleration ##a > g## while in contact with the spring. Since the spring exerts the greatest force when it is the most compressed, the apparatus will undergo the greatest acceleration at that point. So Newton's second law gives

$$\sum F = ma$$

$$kx-mg = ma$$

Therefore, the force constant of the spring should be $$k = \frac{m(a+g)}{x}$$.

b) There are only conservative forces in the system, so energy is conserved. Let point 1 (with ##y=0##) be the point where the apparatus is released and let point 2 be the point where it reaches height ##h##. Then

$$U_1=U_2$$

$$\frac{1}{2}kx^2 = mgh$$

Expressing ##x##

$$x=\sqrt{\frac{2mgh}{k}}$$

However, the

**official solution**at the back of the book isa)

$$k = \frac{m(g+a)^2}{2gh} $$

b)

$$x = \frac{2gh}{g+a} $$

I could 'reverse-engineer' both solutions. However, I don't understand how I should have known to express ##k## in terms of ##m##, ##a##, ##g## and ##h##, and not ##x##. Is it because of part b? Because essentially, both ##k## and ##x## are target variables and only the rest are known?