- #1

lingo

- 4

- 2

- Homework Statement
- A Spring launched cannon consists of a massless platform atop a vertical spring of elastic constant ##k## (top left). The spring must be compressed a distance ##h## (from its neutral position) to engage a latch mechanism that holds the platform in place until launch.

A block of mass m is gently placed on the platform, but the spring only compresses a distance ##h/3## before coming to equilibrium, and thus cannot lock into place (top right). To fully compress the spring and lock the mechanism, you must therefore push the block down with an external applied force of magnitude ##P##. Assume that you start pushing the moment you place the block (bottom left) and continue with a constant force until the latch engages (bottom right).

What is the smallest value for P that will allow the lock to engage? Express your answer as a multiple of the block's true weight, ##mg##.

Hint: You may assume the block ends at rest, at the moment the latch engages.

- Relevant Equations
- ##\Delta E = W_{external}##

##F_s = kx##

So, my question is pertaining more to a specific part of this problem than actually calculating ##P## which I get to be ##P = \frac{kh}{2} - mg##. But I need ##P## in terms of a multiple of ##mg## so I need to find ##k##.

The solution attached uses the fact that when the object comes to a rest on the spring and is not moving, the downward force acting on the block, ##mg##, is equal to the spring force at that compression distance ##F_s = \frac{kh}{3}##. Then ##k = \frac{3mg}{h}##.

However, my solution involves the conservation of energy. Isn't the amount of work done by only the gravitational force, ##mg##, without the involvement of the external force ##P##, equal to the amount of potential energy stored in the spring. So then ##mg(\frac{h}{3}) = \frac{1}{2}k(\frac{h}{3})^2##.

Then solving for ##k## gives ##k = \frac{6mg}{h}##.

I dont know which ##k## is correct; is my approach incorrect?

The solution attached uses the fact that when the object comes to a rest on the spring and is not moving, the downward force acting on the block, ##mg##, is equal to the spring force at that compression distance ##F_s = \frac{kh}{3}##. Then ##k = \frac{3mg}{h}##.

However, my solution involves the conservation of energy. Isn't the amount of work done by only the gravitational force, ##mg##, without the involvement of the external force ##P##, equal to the amount of potential energy stored in the spring. So then ##mg(\frac{h}{3}) = \frac{1}{2}k(\frac{h}{3})^2##.

Then solving for ##k## gives ##k = \frac{6mg}{h}##.

I dont know which ##k## is correct; is my approach incorrect?