A 11.3 weather rocket generates a thrust of 240 . The rocket, pointing upward, is clamped to thetop of a vertical spring. The bottom of the spring, whose springconstant is 570 , is anchored to the ground.

Question

Initially, before the engine is ignited, therocket sits at rest on top of the spring. How much is the springcompressed? << I got 19.4 cm

After the engine is ignited, what is therocket's speed when the spring has stretched 39.0 ?

I tried:

V= √[ 2( 570(0.19)2 + 240 x 0.58 - 11.3 x 9.81x 0.58 )/11.3]

But this is incorrect

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

a= F/m = (Ft - mg )/m = (240 -11.3 x 9.81)/11.3

V=√(2ha) =√(2x 0.39 x 14.1)

Also incorrect.

I tried it a few different ways and I can't figure out what I'm doing wrong. Please help. Thanks.

Answer 1

yes, down .195 meters at start

up .39 meters at end

so moves up .39 + .195 = .585 m

work done by engine = 240*.585 = 140.4 J

so increase in potential energy due to gravity = m g h = 11.3*9.8*.585 = 64.8 J

decrease in potential energy as spring expands from -.195 to 0 = .5*570*.195^2
= 10.8 J

increase in potential energy due to compression of spring from 0 to .39 =
.5 * 570 * .39*2 = 43.3 J

Ke at end = .5 * 11.3 v^2 = 5.65 v^2

work done by thrust + decrease in potential energy = increase in potential energy + 5.65 v^2

140.4 + 10.8 = 64.8 + 43.3 + 5.65 v^2
solve for v

for the second part leave out the decrease and increase of potential energy due to the spring and repeat.

Answer 2

You made the same mistake as in the other problem in assuming the spring force was constant and therefore a due to the spring was constant
in fact the average force is half the maximum for a spring compression or stretch
F average from 0 to x = (1/2) k x

energy is not k x * x but (1/2)kx*x =(1/2)kx^2