Question

A 10.3 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 400 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?
After the engine is ignited, what is the rocket's speed when the spring has stretched 30.0 cm?
For comparison, what would be the rocket's speed after traveling this distance if it weren't attached to the spring?

Answer 1

Answer:

(a) x = 0.25 m

(b) v = 1.46 m/s

(c) v = 2.4 m/s  

Explanation:

mass (m) = 10.3 kg

force from thrust (F) = 240 N

spring constant (k) = 400 N/m

stretch distance from thrust (y) = 30 cm = 0.3 m

acceleration due to gravity (g) = 9.8 m/s^{2}

(A) from mg = kx

compression (x) = mg/ k

x = $\frac{10.3 x 9.8}{400}$

x = 0.25 m

   

(B) from the conservation of forces

  (Fy) + (0.5k$x^{2}$) =  (0.5k$y^{2}$) + mgh + (0.5m$v^{2}$)

v = $\sqrt{[tex]\frac{(Fy) + (0.5k[tex]x^{2}$) -  (0.5k$y^{2}$) - mgh }{0.5m}[/tex]}[/tex]

v =  $\sqrt{[tex]\frac{(240 x 0.3) + (0.5 x 400 x [tex]0.25^{2}$) -  (0.5 x 400 x $0.3^{2}$) - (10.3 x 9.8 x (0.25 + 0.3)) }{0.5 x 10.3}[/tex]}[/tex]

v = 1.46 m/s

                 

(C) if the rocket weren't attached to the spring, the conservation of energy equation becomes

 (Fy) + (0.5k$x^{2}$) = mgh + (0.5m$v^{2}$)

v = $\sqrt{[tex]\frac{(Fy) + (0.5k[tex]x^{2}$)  - mgh }{0.5m}[/tex]}[/tex]

v =  $\sqrt{[tex]\frac{(240 x 0.3) + (0.5 x 400 x [tex]0.25^{2}$)  - (10.3 x 9.8 x (0.25 + 0.3)) }{0.5 x 10.3}[/tex]}[/tex]

v = 2.4 m/s