Question

1. A 12 weather rocket generates a thrust of 200 . The rocket, pointing upward, is clamped to the top of a vertical spring, 550 / . The spring is anchored to the ground. a. Initially, before the engine is ignited, the rocket sits at rest on the 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 40 ?

Answer 1

a) 0.214 m (21.4 cm)

b) 1.8 m/s

Explanation:

a)

When the rocket sits on the spring, the weight of the rocket (acting downward) is balanced by the restoring force on the spring (upward).

So, we can write:

$mg=kx$

where:

- The term on the left is the weight of the rocket, with

m = 12 kg is its mass

$g=9.8 m/s^2$ is the acceleration due to gravity

- The term on the right is the restoring force, where

k = 550 N/m is the spring constant

x is the compression of the spring

And solving for x, we find:

$x=\frac{mg}{k}=\frac{(12)(9.8)}{550}=0.214 m$

Which corresponds to 21.4 cm.

b)

We can solve this part by using the law of conservation of energy.

At the beginning, when the engine is off, the rocket is at rest sitting on the spring; the total energy of the system therefore is the elastic potential energy stored in the spring, which is:

$E=\frac{1}{2}kx^2=\frac{1}{2}(550)(0.214)^2=12.6 J$

Then, when the engine is turned on, the rocket stars moving upward. The string will move from a position of x = -21.4 cm (compression) to a position of x = +40 cm (0.40 m, stretching), so the new elastic potential energy of the spring will be

$E'=\frac{1}{2}kx'^2 = \frac{1}{2}(550)(0.40)^2=44 J$

Moreover, the work done by the thurst of the rocket, of magnitude F = 200 N upward, is equal to the magnitude of F time the displacement of the rocket, so

$W=F(x'-x)=(200)(0.40-(-0.214))=122.8 J$

Also, the increase in gravitational potential energy of the rocket is

$\Delta PE=mg(x'-x)=(12)(9.8)(0.40-(0.214))=72.2 J$

So, according to the law of conservation of energy, the final kinetic energy of the rocket will be:

$KE=E-E'+W-\Delta PE=12.6-44+122.8-72.2=19.2 J$

The kinetic energy can be rewritten as

$KE=\frac{1}{2}mv^2$

where m = 12 kg is the mass of the rocket; and solving for v, we find

$v=\sqrt{\frac{2KE}{m}}=\sqrt{\frac{2(19.2)}{12}}=1.8 m/s$