A 12.0 kg weather rocket generates a thrust of 210 N . The rocket, pointing upwa

Question

A 12.0 kg weather rocket generates a thrust of 210 N . The rocket, pointing upward, is clamped to the top of a vertical spring. The bottom of the spring, whose spring constant is 490 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 40.0 cm ? c)For comparison, what would be the rocket's speed after traveling this distance if it weren't attached to the spring?

Explanation / Answer

a)  F = kx. You know the force compressing it, you know the spring constant

k= 490 N/m , F= 210 N

spring compression (x)= F/k =210/490 --> x = 0.42857

b) xi = 0.42857 , xf = 40cm = 0.40 m , mass (M) =21 kg

initial potential engergy =PEi(spring) =1/2* k * xi^2 = 0.5(490)(0.42857^2) = 44.9997 = 45 J

final potential engergy = PEf(spring) = 0.5(490)(0.40^2) = 39.2 J

initial potential engergy= PEi(gravity) = 0
final potential engergy = PEf(gravity) = M*g*(xi + xf) = 12*(9.8)(0.42857 + 0.4) = 97.439 J
work done(W) =force * displacement = 210(0.42857 + 0.4) = 173.9997 =174 J

PE(spring) + KE + PE(gravity) = W
(39.2-45) + 0.5(12)(v²) + (97.439 - 0) = 174
- 5.8 + 6v² + 97.439 = 120
6v² = 82.361
v² = 13.72
v = 3.704m/s

c) if it's not attached to the spring then PEf(spring) = 0
PE(spring) + KE + PE(gravity) = W
(0 - 45) + 0.5(12)v² + 97.439= 174
6v² = 121.564
v² = 20.26
v = 4.5 m/s

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