Rocket Thrust From a Spring (Work)

In summary, we are trying to find the velocity of a rocket attached to a spring after it has traveled a certain distance. We can calculate the velocity by using the equation KE = 1/2mv^2 and taking into account the contributions from the energy of the compressed spring, the force of the rocket engine, and the force of gravity.
  • #1
sam2k2002
10
0

Homework Statement


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.

Homework Equations



Ug = mgh
Us = 1/2 k (delta_x)^2
KE = 1/2mv^2
W_thrust = F_thrust*distance

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

Homework Statement


Homework Equations


The Attempt at a Solution

 
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  • #2
I think the answer is 3.79 m/s
 
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  • #3
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?
 
  • #4
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.
 
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  • #5
Nope.. That was my last attempt, the answer turned out to be 3.67. Any idea what we're doing wrong?
 
  • #6
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
 
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  • #7
DocZaius said:
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 which the solution is x = (1/2) a t^2
 
  • #8
sam2k2002 said:
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.
 
  • #9
sam2k2002 said:
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%
 
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  • #10
DocZaius said:
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, which comes from.
- The energy of the compressed spring.
- The force of the rocket engine which acts over a 0.535 m stretch.
- The force of gravity which 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.
 

1. What is the concept behind rocket thrust from a spring?

The concept behind rocket thrust from a spring is based on the principle of conservation of energy. When a spring is compressed, it stores potential energy. When released, this potential energy is converted into kinetic energy, causing the spring to expand and produce thrust.

2. How does the amount of work done on the spring affect the rocket thrust?

The amount of work done on the spring directly affects the rocket thrust. According to the work-energy theorem, the work done on an object is equal to the change in its kinetic energy. Therefore, the more work done on the spring, the greater the change in its kinetic energy and the greater the resulting thrust.

3. Can rocket thrust from a spring be used to propel a rocket into space?

No, rocket thrust from a spring alone is not sufficient to propel a rocket into space. While it can generate a burst of thrust, it is not sustainable and cannot provide the continuous acceleration needed to overcome Earth's gravity and reach space.

4. How does the spring constant affect the rocket thrust?

The spring constant, which is a measure of the stiffness of the spring, affects the rocket thrust in two ways. Firstly, a higher spring constant means that more force is required to compress the spring, resulting in higher potential energy and therefore a greater amount of thrust. Secondly, a higher spring constant also means that the spring will return to its original length faster, resulting in a shorter duration of thrust.

5. Are there any other factors that can affect the rocket thrust from a spring?

Yes, there are several other factors that can affect the rocket thrust from a spring. These include the initial compression of the spring, the mass of the rocket, and the efficiency of the conversion of potential energy into kinetic energy. Additionally, external factors such as air resistance can also impact the resulting thrust.

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