Question

A 11.7kg weather rocket generates a thrust of 200N . The rocket, pointing upward, is clamped to the top of a vertical spring. The bottom of the spring, whose spring constant is 550N/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 32.0cm ?

C)For comparison, what would be the rocket's speed after traveling this distance if it weren't attached to the spring?

Answer 1

1)
m*g = k*x
x = (m*g)/k
x = (11.7kg * 9.81m/s^2)/(550N/m)
x = 0.208685175 m

x_1=0.208685175
lets call this x_1



2)
F_thrust*y + 1/2*k*(x_1)^2 = 1/2*k*(x_2)^2 + m*g*y + 1/2*m*v^2

where
F_thrust=210N
y=(x_1+x_2)
k=550N/m
x_1= 0.208685175 m (this is from part 1)
x_2= 0.32m (this is the distance in the question in part 2)
m=11.7kg
g=9.81m/s^2

F_thrust*y + 1/2*k*(x_1)^2 = 1/2*k*(x_2)^2 + m*g*y + 1/2*m*v^2

200*(0.2086851+0.32) + 1/2*550*(.0.208685175 )^2 = 1/2*550*(0.32)^2 + 11.7*9.81*(0.208685175 +0.31) + 1/2*11.7*v^2

120.12 + 13.7288 = 19.22 + 60.04 + 5.35*v^2
150.2339=5.85*v^2

v^2=25.6810

v=5.0676 m/s



3) you just take out the 1/2*k*(x_2)^2 term since the spring is not attached anymore and solve like 2)

F_thrust*y + 1/2*k*(x_1)^2 = m*g*y + 1/2*m*v^2
200*(0.2086851+0.32) + 1/2*550*(.0.208685175 )^2 = 1/2*550*(0.32)^2 + 1/2*11.7*v^2

V=3.9125 m/s