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alphaQ

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## Homework Statement

if you wanted to build a spring launched cannon that will shoot you over a building that is 35 m high and 30 m wide, and the cannon is being shot at 60 degrees. If the cannon can be no more than 2 m long, what spring constant do you need in the spring to make this work? here is what I have been able to do, any feed back would be appreciated.

## Homework Equations

^2= r/()PE = KE

**½ Fx = ½ mv^2**

K = F / x

## The Attempt at a Solution

to find the velocity I used:

v = sqrt((30* 9.8 m/s^2)/sin(60))

v = 18.425 m/s

Theoretically, humans can handle any velocity, it’s acceleration that we cannot handle well. A typical human can handle up to 5Gs of acceleration, that’s 49 ^2.

Since the maximum PE of the spring is the maximum KE of the mass PE = KE ½ Fx = ½ mv^2 Rearranging the formula, we get: = ^2/

F = ma = (70kg) (49m/s2) = 3430 N

KE = ½ mv2 = ½ (70kg) (18.425 m/s2) = 644.875 J

So, x = (644.875 J) / (½ ) (3430 N) x = 0.3760204082 m

We can now use x to solve for the spring constant

K = F / x

K = 3430N / 0.3760204082m + 2m(for the length of the cannon)

K = 1443.590294 N/m

is this K the right spring constant for this question?

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