How Does a Spring Affect the Motion of a Rocket?

[shrutij]

Homework Statement

A 10.1 kg weather rocket generates a thrust of 193.0 N. The rocket, pointing upward, is clamped to the top of a vertical spring. The bottom of the spring, whose spring constant is 415.0 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?
(b) After the engine is ignited, what is the rocket's speed when the spring has stretched 12.5 cm past its natural length?
(c)What would be the rockets speed after traveling the distance if it weren't tied down to the spring?

Homework Equations

Usp= 1/2*k*x^2

The Attempt at a Solution

I'm not too sure how to even begin thinking about this question.
For (a) Am I right in the assumption that the weight of the rocket is directly related to the compression of the spring, thus the gravitational potential energy of the rocket should be equal to the spring energy?
I tried mgh=1/2kx^2, where h=x. From this I got x=2(mg)/k, which is not the right answer.
How would I break down this question? Any help is appreciated!

 

[housemartin]

Hi ho!
for a) Your assumption about equal potential energies is simply wrong (you can always add a constant to potential energy). What matters is difference in potential energy, not a numerical value at certain point in space. Since it is said that rocket is at rest, you can be sure that the total force acting on it is zero. There are only two forces in this situation i can think of. One of them depends on spring compression.
b) here you can either solve Newton's equation of motion, or use work energy theorem.
c) same as b)

 

[shrutij]

I'm not sure what you mean when you say two forces
One is the weight of the rocket, but the spring compression is the energy stored in the spring that will get converted to the kinetic energy of rocket, right? How do i relate the fnet=0 to the spring compression?

 

[shrutij]

Can someone please help me with this question? Really having trouble understanding the concepts.
Thanks!

 

[asteriatic]

To begin, let's first consider the forces acting on the rocket before the engine is ignited. We have the weight of the rocket acting downwards and the force of the spring acting upwards. Since the rocket is at rest, these two forces must be balanced. Therefore, we can say that:

Weight of rocket = Force of spring

Now, let's use the equation for the force of a spring, F = -kx, where k is the spring constant and x is the displacement from the equilibrium position. Since the rocket is clamped to the top of the spring, the displacement is equal to the compression of the spring. So, we can rewrite the equation as:

Weight of rocket = kx

Substituting in the given values, we get:

(10.1 kg)(9.8 m/s^2) = (415.0 N/m)(x)

Solving for x gives us the answer for (a) as x = 0.242 m.

Moving on to part (b), we can use the work-energy theorem to find the speed of the rocket when the spring has stretched 12.5 cm past its natural length. We know that the work done by the spring is equal to the change in kinetic energy of the rocket. So, we can write:

Work done by spring = Change in kinetic energy

The work done by the spring is given by the formula Usp = 1/2 kx^2, where Usp is the potential energy stored in the spring. So, we can rewrite the equation as:

1/2 kx^2 = 1/2 mv^2

Substituting in the given values and solving for v gives us the answer for (b) as v = 24.7 m/s.

Finally, for part (c), we can use the conservation of energy principle to find the rocket's speed after it has traveled the distance if it weren't tied down to the spring. At the point where the spring has stretched 12.5 cm, all of the potential energy stored in the spring has been converted into kinetic energy of the rocket. So, we can write:

1/2 kx^2 = 1/2 mv^2

Substituting in the given values and solving for v gives us the answer for (c) as v = 49.4 m/s. This is the maximum speed the rocket would reach if it weren't tied down to