Can a Compressed Spring Launch You Over a 15-ft Wall?

[yellowstriker]

There is a "cannon" that is a 3-ft diameter tube that is 12-ft. long with a stiff spring inside. The spring will be compressed to 1/10 its normal length and held. You will be on the spring. With perfect aim you will fly through the air over a 15-ft wall and land safely in a net. The spring is 10-ft. long. When you hang on the spring w/o touvhing the ground it stretches by 2-ft. Is it possible to make it over the wall?

 

[HallsofIvy]

This smells like homework!

I don't understand the problem. YOu state "you will fly through the air over a 15-ft wall and land safely in a net", then ask "is it possible to make it over the wall". Didn't you just say it was??

I guess the problem is to find the potential energy stored in the spring when it is compressed, convert that to kinetic energy as you leave the cannon to find your initial speed and then use kinematics to determine if the maximum point of the trajectory really is more than 15 ft. To do that, you need to know the "spring constant" for the spring.

That's probably the point of "When you hang on the spring w/o touvhing the ground it stretches by 2-ft." but, while I know my weight and so could find the force on the spring when I hang on it I don't know yours so my solution wouldn't necessarily work for you! Does the problem give a weight for the person?

 

[Andrew Mason]

There is a "cannon" that is a 3-ft diameter tube that is 12-ft. long with a stiff spring inside. The spring will be compressed to 1/10 its normal length and held. You will be on the spring. With perfect aim you will fly through the air over a 15-ft wall and land safely in a net. The spring is 10-ft. long. When you hang on the spring w/o touvhing the ground it stretches by 2-ft. Is it possible to make it over the wall?

The condition for success is:

Spring potential energy > mgh + KE_top where h is the height of the wall and KE_top is your kinetic energy at maximum height (vertical speed = 0). There has to be some extra energy in the spring to provide horizontal acceleration to give you a horizontal velocity to actually get over the wall but ignore that for the moment

So:
$$\frac{1}{2}kD^2 > mgh$$

To determine K, we know that it stretches 2 feet due to your weight so:

kd=mg where d =2 so k = mg/2

I think you will find that the inequality is satisfied leaving more than enough for the needed Kinetic energy to provide horizontal travel.

AM