ttps://www.chegg.com/studylqa/post Search.. MasteringPhysics Work and Kinetic En

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ttps://www.chegg.com/studylqa/post Search.. MasteringPhysics Work and Kinetic Energy . Google Chrome Secure https /session mastenngphysics com/myct/itemView?assignmentProblemID-87499471 he Problem 9.70 previous 27 of 27 return to assign Problem 9.70 Part B After the engine is ignited, what is the rocket's speed when the spring has stretched 400 cm Express your answer with the appropriate units. A 10 2 kg weather rocket generates a thrust of 200 The rocket, pointing upward, is clamped to the top of a vertical spring. The bottom of the spring whose spring constant is 460 N/m, is anchored to the ground V1= 265 Subrnit MY Answers Give Up All attempts used; correct answer displayed Part c For comparison, what would be the rocker's speed after traveling this distance if it weren't attached to the spring? Express your answer with the appropriate units. 0 U2 = I l'alue Units T CH Submit My Answers Give Up e here to search

Explanation / Answer

1) here m= mass, k= spring constant, x= position!

m*g = k*x

x = (m*g)/k

x = (10.2kg * 9.81m/s^2)/(460N/m)

x = 0.217526 m

x = 0.2175m

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=200N

y=(x_1+x_2)

k=460N/m

x_1= 0.2175m (this is from part 1)

x_2= 0.4m (this is the distance in the question in part 2)

m=10.2kg

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*(0.2175+0.4) + 1/2*460*(.2175)^2 = 1/2*460*(0.4)^2 + 10.2*9.81*(0.2175+0.4) + 1/2*10.2*v^2

=> v = 2.64916 m/s

~2.65 m/s

Now we need to 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*(V2)^2

=> V2= 3.77276 m/s

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