Calculating spring constant of a spring loaded cannon

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Homework Help Overview

The discussion revolves around calculating the spring constant for a spring-loaded cannon designed to launch a person over a building that is 35 m high and 30 m wide, with a launch angle of 60 degrees. The cannon's length is limited to 2 m.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to calculate the necessary spring constant using energy conservation principles and projectile motion equations. Some participants question how to incorporate both the height and width of the building into the calculations. Others suggest focusing on the minimum energy required to clear the building and emphasize the importance of trajectory analysis.

Discussion Status

The discussion is ongoing, with participants exploring different aspects of the problem. Some guidance has been offered regarding the need to determine the minimum speed required to clear the building and the trajectory equations necessary for that calculation. There is no explicit consensus on the approach yet.

Contextual Notes

Participants are grappling with how to integrate the dimensions of the building into their calculations and are seeking clarification on relevant equations, particularly those related to projectile motion.

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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alphaQ said:
v = sqrt((30* 9.8 m/s^2)/sin(60))
Not so fast: the building has a height and a width. You must clear it completely.
Draw a diagram.
 
haruspex said:
Not so fast: the building has a height and a width. You must clear it completely.
Draw a diagram.
thats where i have the problem, i don't know how to incorporate the width and height into the equation
 
alphaQ said:
thats where i have the problem, i don't know how to incorporate the width and height into the equation
You are looking for the minimum energy, so only just clearing the building.
If the launch is at speed v and distance x from the building, what equations can you write for just clearing it?
(Take the launch speed as being at ground level.)
 
haruspex said:
You are looking for the minimum energy, so only just clearing the building.
If the launch is at speed v and distance x from the building, what equations can you write for just clearing it?
(Take the launch speed as being at ground level.)
i was thinking of getting the spring constant and then checking whether that constant would allow me to get over the height of the building
 
alphaQ said:
i was thinking of getting the spring constant and then checking whether that constant would allow me to get over the height of the building
To find the spring constant you need the launch energy, and to find that you need to find the minimum speed that will get you over the building. Start with solving the trajectory.
 
haruspex said:
To find the spring constant you need the launch energy, and to find that you need to find the minimum speed that will get you over the building. Start with solving the trajectory.
can you please give me an equation to get the launch energy?
 
alphaQ said:
can you please give me an equation to get the launch energy?
As I wrote, you need to start by finding the trajectory.
You did not quote any relevant equations for that. Do you know what I mean by SUVAT equations? If not, look them up.
 

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