# Rocket Thrust From a Spring (Work)

by sam2k2002
Tags: rocket, spring, thrust, work
 P: 10 1. The problem statement, all variables and given/known data A 11.7 kg weather rocket generates a thrust of 240 N. The rocket, pointing upward, is clamped to the top of a vertical spring. The bottom of the spring, whose spring constant is 560 N/m, is anchored to the ground. (a) Initially, before the engine is ignited, the rocket sits at rest on top of the spring. How much is the spring compressed? 20.5cm (.205m) (b) After the engine is ignited, what is the rocket's speed when the spring has stretched 33.0 cm? 2.87m/s (c) For comparison, what would be the rocket's speed after traveling this distance if it weren't attached to the spring? This is the question I have a problem with. 2. Relevant equations Ug = mgh Us = 1/2 k (delta_x)^2 KE = 1/2mv^2 W_thrust = F_thrust*distance 3. The attempt at a solution Ok, so for Part (b) I used the following equation to find the velocity: W_thrust + U_s(using x_1) = U_s(using x_2) + U_g + KE I was able to solve for the v in KE correctly using this equation. So in order to find the velocity of the rocket when the spring is taken out, I take the U_s out of the equation: W_thrust = U_g + KE Where the distance here will be x_1 + x_2. (x_compressed + x_stretched = x_total) I've done this problem a number of ways.. work, energy, kinematics.. and i keep arriving at the (incorrect) answer v = 3.38m/s. Can anyone help me spot my error? Thanks in advance. -Sam 1. The problem statement, all variables and given/known data 2. Relevant equations 3. The attempt at a solution
 P: 25 I think the answer is 3.79 m/s
 P: 10 Nope, that doesn't work. Here's what i have for the energy equation. Work_thrust = U_g + KE (240N)(.535m) = (11.7kg)(9.81)(.535m) + 1/2(11.7kg)v^2 With that equation I get 3.38m/s, which is wrong. Maybe I'm using the wrong height? The problem asks for the velocity at the height of the previous question, where x1 was -.205m and x2 was +.33m, so the deltaX would just be .535, right?
P: 25

## Rocket Thrust From a Spring (Work)

I think I figured out the problem, one sec....

Ok, I get 3.79 m/s. Is this correct?

If it is, then I am pretty sure that I know what you did wrong and why all the other methods you used gave you the same answer.
 P: 10 Nope.. That was my last attempt, the answer turned out to be 3.67. Any idea what we're doing wrong?
 P: 281 I thought thrust was a force. Knowing force and mass, we know the rocket's acceleration. Knowing that, we know velocity at any given distance. Where am I going wrong? f=m*a (240)=(11.7)*a a=20.5128 dx/dt^2=a dx/dt^2=20.5128 0.33/dt^2=20.5128 dt^2=0.0161 dt=0.1268 dx/dt=2.6018 m/s
P: 454
 Quote by DocZaius I thought thrust was a force. Knowing force and mass, we know the rocket's acceleration. Knowing that, we know velocity at any given distance. Where am I going wrong? f=m*a (240)=(11.7)*a a=20.5128 dx/dt^2=a dx/dt^2=20.5128 0.33/dt^2=20.5128 dt^2=0.0161 dt=0.1268 dx/dt=2.6018 m/s
You're forgetting gravity.

Nothing after dx/dt^2=a makes sense. This is a second order differential
equation, of wich the solution is x = (1/2) a t^2
P: 454
 Quote by sam2k2002 Nope.. That was my last attempt, the answer turned out to be 3.67. Any idea what we're doing wrong?
I think the spring will still push against the rocket, so you can add the energy of
the compressed spring to the rocket.
P: 281
 Quote by sam2k2002 Nope.. That was my last attempt, the answer turned out to be 3.67. Any idea what we're doing wrong?
I just came within 0.262% of that number

3 components of force going on here:

1) the rocket's thrust
2) the spring's push (as kamerling says, the spring is still pushing)
3) the earth's pull

1)
acceleration from rocket thrust

f=ma
240=11.7a
a=20.5128

2)
acceleration from spring's push

f=-kx
f=-(-0.205)(560)
f=114.8

f=ma
114.8=(11.7)a
a=9.812

3)
the earth's pull

a=-9.8

Now add them all up and you get
a=20.5248

x(t)=(1/2)a*t^2
0.33=(1/2)(20.5248)*t^2
t=0.1793

v(t)=at
v=20.5248*.1793
v=3.68
error margin of 0.272%
P: 454
 Quote by DocZaius I just came within 0.262% of that number 3 components of force going on here: 1) the rocket's thrust 2) the spring's push (as kamerling says, the spring is still pushing) 3) the earth's pull 1) acceleration from rocket thrust f=ma 240=11.7a a=20.5128 2) acceleration from spring's push f=-kx f=-(-0.205)(560) f=114.8 f=ma 114.8=(11.7)a a=9.812 3) the earth's pull a=-9.8 Now add them all up and you get a=20.5248 x(t)=(1/2)a*t^2 0.33=(1/2)(20.5248)*t^2 t=0.1793 v(t)=at v=20.5248*.1793 v=3.68 error margin of 0.272%
The acceleration of the spring isn't constant. The acceleration of the spring only works from x= -20.5 up to x=0. You only computed x and v when x>=0, but the initial position of the rocket is at -20.5.

Because of the non-constant acceleration of the spring it's rather hard to compute x(t) or v(t) and it's easier to compute the kinetic energy of the rocket, wich comes from.
- The energy of the compressed spring.
- The force of the rocket engine wich acts over a 0.535 m stretch.
- The force of gravity wich also acts over a 0.535 m stretch, but in the opposite direction of the velocity, so its contribution is negative.

If you add these up, and use (kinetic energy) = (1/2)mv^2 you get the right answer.

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