Math Problem Statement

A 10.9 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 480 N/m , is anchored to the ground. After the engine is ignited, what is the rocket's speed when the spring has stretched 37.0 cm ?

Solution

To find the rocket's speed when the spring has stretched 37.0 cm, let's go through the following steps:

  1. Identify the forces and energy involved:

    • The rocket has a mass m=10.9kgm = 10.9 \, \text{kg}.
    • The gravitational force acting on the rocket is Fgravity=mgF_{\text{gravity}} = mg, where g=9.8m/s2g = 9.8 \, \text{m/s}^2.
    • The thrust from the rocket's engine is Fthrust=210NF_{\text{thrust}} = 210 \, \text{N}.
    • The spring has a spring constant k=480N/mk = 480 \, \text{N/m} and stretches by x=37.0cm=0.37mx = 37.0 \, \text{cm} = 0.37 \, \text{m}.
  2. Calculate the net force exerted on the rocket:

    • The forces acting on the rocket in the vertical direction are:

      • Upward thrust, Fthrust=210NF_{\text{thrust}} = 210 \, \text{N}
      • Downward gravitational force, Fgravity=mg=10.9×9.8=106.82NF_{\text{gravity}} = mg = 10.9 \times 9.8 = 106.82 \, \text{N}
      • Downward spring force, Fspring=kx=480×0.37=177.6NF_{\text{spring}} = kx = 480 \times 0.37 = 177.6 \, \text{N}
    • The net force FnetF_{\text{net}} is given by: Fnet=Fthrust(Fgravity+Fspring)F_{\text{net}} = F_{\text{thrust}} - (F_{\text{gravity}} + F_{\text{spring}}) Fnet=210(106.82+177.6)=210284.42=74.42NF_{\text{net}} = 210 - (106.82 + 177.6) = 210 - 284.42 = -74.42 \, \text{N}

    This negative value indicates a downward net force on the rocket.

  3. Use energy conservation to find the speed:

    • We apply the work-energy theorem, which states: ΔKE=Wnet\Delta KE = W_{\text{net}} where ΔKE\Delta KE is the change in kinetic energy and WnetW_{\text{net}} is the work done by the net force.

    • The work done by the net force as the spring stretches by x=0.37mx = 0.37 \, \text{m} is: Wnet=Fnetx=74.42×0.37=27.53JW_{\text{net}} = F_{\text{net}} \cdot x = -74.42 \times 0.37 = -27.53 \, \text{J}

    • Assuming the rocket started from rest, the initial kinetic energy KEinitial=0KE_{\text{initial}} = 0. So, ΔKE=KEfinalKEinitial=KEfinal=27.53J\Delta KE = KE_{\text{final}} - KE_{\text{initial}} = KE_{\text{final}} = -27.53 \, \text{J}

    Since kinetic energy cannot be negative, this value suggests the rocket doesn't actually gain enough energy to move upward under the given conditions. Hence, the rocket remains stationary, with downward forces exceeding the thrust provided.

Would you like more details or have any questions?

Here are some further questions for exploration:

  1. What would happen if the thrust were stronger than the combined gravitational and spring forces?
  2. How does increasing the spring constant impact the rocket's motion?
  3. How would a lighter rocket impact the calculations?
  4. What if the spring were attached horizontally instead of vertically?
  5. Could we calculate the maximum stretch of the spring before the rocket starts to ascend?

Tip: In cases involving both forces and energy, checking both approaches helps ensure the solution’s accuracy.

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Math Problem Analysis

Mathematical Concepts

Physics (Forces)
Energy Conservation
Kinetics

Formulas

F_gravity = mg
F_spring = kx
F_net = F_thrust - (F_gravity + F_spring)
Work-energy theorem: ΔKE = W_net

Theorems

Work-Energy Theorem

Suitable Grade Level

Grades 11-12