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. Before the engine is ignited, the rocket sits on a spring which is compressed Δy = 22.3 cm After the engine is ignited, the rocket's speed when the spring has stretched 37.0 cm is 2.71 m/s. For comparison, what would be the rocket's speed after traveling this distance if it weren't attached to the spring? Use the same approach as in the previous part but note that the spring is not included in the final state of the system. Do not round until the final step

Solution

To solve this problem, let's analyze the forces acting on the rocket and use the work-energy principle. The problem asks us to find the rocket's speed after traveling 37.0 cm (0.37 m) upward without the influence of the spring.

Given Data

  • Mass of the rocket, m=10.9kgm = 10.9 \, \text{kg}
  • Thrust force generated by the rocket, Fthrust=210NF_{\text{thrust}} = 210 \, \text{N}
  • Gravitational force acting on the rocket, Fgravity=mg=10.9×9.81m/s2106.8NF_{\text{gravity}} = mg = 10.9 \times 9.81 \, \text{m/s}^2 \approx 106.8 \, \text{N}
  • Distance traveled upward, d=0.37md = 0.37 \, \text{m}
  • Initial speed, v0=0m/sv_0 = 0 \, \text{m/s} (we'll assume it starts from rest)

Steps to Solution

  1. Calculate the Net Force Acting on the Rocket (Without the Spring):

    The net force acting on the rocket is the difference between the thrust and the gravitational force: Fnet=FthrustFgravity=210N106.8N=103.2NF_{\text{net}} = F_{\text{thrust}} - F_{\text{gravity}} = 210 \, \text{N} - 106.8 \, \text{N} = 103.2 \, \text{N}

  2. Work Done by the Net Force:

    The work done by the net force over the distance dd is: W=Fnet×d=103.2N×0.37m=38.184JW = F_{\text{net}} \times d = 103.2 \, \text{N} \times 0.37 \, \text{m} = 38.184 \, \text{J}

  3. Use the Work-Energy Theorem:

    According to the work-energy theorem, the work done on the rocket is equal to the change in its kinetic energy: W=ΔKE=12mv212mv02W = \Delta KE = \frac{1}{2}mv^2 - \frac{1}{2}mv_0^2 Since the initial speed v0=0v_0 = 0, the equation simplifies to: W=12mv2W = \frac{1}{2}mv^2

  4. Solve for vv:

    Substitute the values and solve for vv: 38.184=12×10.9×v238.184 = \frac{1}{2} \times 10.9 \times v^2 v2=38.184×210.9=76.36810.97.004v^2 = \frac{38.184 \times 2}{10.9} = \frac{76.368}{10.9} \approx 7.004 v7.0042.65m/sv \approx \sqrt{7.004} \approx 2.65 \, \text{m/s}

Final Answer

If the rocket weren't attached to the spring, its speed after traveling 37.0 cm would be approximately: 2.65m/s\boxed{2.65 \, \text{m/s}}

Would you like further details, or do you have any questions?


Here are five related questions to consider:

  1. How would the speed change if the thrust force was increased?
  2. What would happen to the rocket's motion if the mass was reduced while keeping the thrust the same?
  3. How would you find the speed if the rocket were moving horizontally instead of vertically?
  4. How does adding or removing external forces, like the spring, affect the work-energy calculations?
  5. Can you calculate the potential energy stored in the spring initially compressed by 22.3 cm?

Tip: When analyzing forces, always separate gravitational forces and other external forces to understand the net effect on motion more clearly.

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Math Problem Analysis

Mathematical Concepts

Physics
Work-Energy Principle
Forces
Kinetic Energy

Formulas

Net force: F_net = F_thrust - F_gravity
Work done: W = F_net × d
Kinetic energy: KE = (1/2)mv^2

Theorems

Work-Energy Theorem

Suitable Grade Level

Grades 11-12