Rocket speed without spring attachment using work-energy theorem

[turtledove]

Homework Statement

A 10.2 kg weather rocket generates a thrust of 177.0 N. The rocket, pointing upward, is clamped to the top of a vertical spring. The bottom of the spring, whose spring constant is 385.0 N/m, is anchored to the ground. Initially, before the engine is ignited, the rocket sits at rest on top of the spring. How much is the spring compressed? (ANS: 0.26 m)

a) After the engine is ignited, what is the rocket's speed when the spring has stretched 12.0 cm past its natural length? (ANS: 2.78 m/s)

b) What would be the rockets speed after traveling the distance if it weren't tied down to the spring? (ANS: ?)

Homework Equations

W = Ef - Ei
Ee = 0.5 kx^2
Ek = 0.5 mv^2
Ep = mgh

The Attempt at a Solution

PART B ONLY:
[Fthrust x d] = [mgh + 0.5mv^2]f - [0]i
[(177)(0.26 + 0.12)] = [(10.2)(9.8)(0.26 + 0.12)] + [0.5(10.2)v^2]
v = 2.40 m/s <-- this is WRONG

I was able to solve PART A, but I don't know what to do when the rocket is not attached to the spring. I suppose it might be Vf^2 = Vi^2 + 2ad, but I don't know what distance to use. In my attempt, I just used the formula I used for PART A, WITHOUT the elastic energy added. Somebody pleaseee help me!

 

[PhanthomJay]

Homework Statement

A 10.2 kg weather rocket generates a thrust of 177.0 N. The rocket, pointing upward, is clamped to the top of a vertical spring. The bottom of the spring, whose spring constant is 385.0 N/m, is anchored to the ground. Initially, before the engine is ignited, the rocket sits at rest on top of the spring. How much is the spring compressed? (ANS: 0.26 m)

a) After the engine is ignited, what is the rocket's speed when the spring has stretched 12.0 cm past its natural length? (ANS: 2.78 m/s)

b) What would be the rockets speed after traveling the distance if it weren't tied down to the spring? (ANS: ?)

Homework Equations

W = Ef - Ei
Ee = 0.5 kx^2
Ek = 0.5 mv^2
Ep = mgh

The Attempt at a Solution

PART B ONLY:
[Fthrust x d] = [mgh + 0.5mv^2]f - [0]i
[(177)(0.26 + 0.12)] = [(10.2)(9.8)(0.26 + 0.12)] + [0.5(10.2)v^2]
v = 2.40 m/s <-- this is WRONG

I was able to solve PART A, but I don't know what to do when the rocket is not attached to the spring. I suppose it might be Vf^2 = Vi^2 + 2ad, but I don't know what distance to use. In my attempt, I just used the formula I used for PART A, WITHOUT the elastic energy added. Somebody pleaseee help me!

Yes, you leave out the final elastic PE of the spring , but the initial elastic PE of the spring is still there, even though the rocket is not attached to it (the spring is initially compressed whether attached or not). Don't leave that initial elastic energy out.

 

[turtledove]

oh ! ok, i was thinking about that too, but i don't know which distances to use? this would be my attempt:

Fthrust x d + 0.5kx^2 = mgh + 0.5mv^2
(177)(0.38) + 0.5(385)(0.26)^2 = (10.2)(9.8)(0.38) + 0.5(10.2)v^2
v = 2.88 m/s

This answer makes sense, because it is greater than the speed with the spring...however, I am not sure if i should use 0.38 as the distance. if you could clear this up for me that would be great!

 

[PhanthomJay]

oh ! ok, i was thinking about that too, but i don't know which distances to use? this would be my attempt:

Fthrust x d + 0.5kx^2 = mgh + 0.5mv^2
(177)(0.38) + 0.5(385)(0.26)^2 = (10.2)(9.8)(0.38) + 0.5(10.2)v^2
v = 2.88 m/s

This answer makes sense, because it is greater than the speed with the spring...however, I am not sure if i should use 0.38 as the distance. if you could clear this up for me that would be great!

Your equation is correct as written. The problem asks for the speed after the rocket has traveled a distance of 0.38 m from its starting point.

 

[turtledove]

awesomee, thanks !