Spring cannon, initial velocity, and compression

[fischelr]

Homework Statement

Part a.) You launch a 0.5 kg water balloon with a spring cannon in order to his someone 20m away. The spring in the cannon has a spring constant of 350 N/m. The spring cannon fires the balloon at a 30 degree angle with respect to the ground. What is the initial velocity necessary for the balloon to travel 20 meters?

Part b.) How much does the spring need to be compressed in order to achieve this initial velocity? Assume that 5 joules of heat is lost as the spring expands.

Homework Equations

F_net = F_spring - lwl = ma
sohcahtoa
*something for energy loss*

The Attempt at a Solution

Part a.) V initial was found to be 14.09 m/s

Part b.)
First I determined the lwl parallel be 30sin 4.9=2.45N

F_net = F_spring - lwl = ma
=k\Deltax - lwl = ma
=350 x - 2.45 = 2.45
350x = 4.90
x=.014m
x-1.4cm


I have found that the spring compression would need to be 1.4cm to launch a .5kg ball 20 meters. However, this sounds small to me. Am I missing a force on the spring? I thought about including power, but I do not have the acceleration of the ball leaving the spring.

Also, I am not sure how to subtract 5 joules of energy due to heat loss.

 

[rock.freak667]

For part a, I did not get 14.09 m/s, can you post how you got that answer?

For part b, you could use conservation of energy here. But resultant force or 'ma' would not be 2.45 N, you'd need to find the acceleration.

 

[twofunky]

On part B would it not simply be the potential energy of the spring = to the kinetic energy plus heat?

so .5(350)x^2 = .5(.5)14.09^2 + 5

x^2 = 54.6/175

x= .56 m

 

[rock.freak667]

Yes that would be correct to do.

 

[TexasBB12]

would a velocity of 15.04 be correct for this?

 

[rock.freak667]

Yes that should be the initial velocity.

 

[stephjuly28]

what equation did you use to get a velocity of 15.04? Thanks!

 

[rock.freak667]

what equation did you use to get a velocity of 15.04? Thanks!

The better question is what did you use to get 14.09 m/s?