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

Solution

To determine the rocket's speed after traveling the same distance if it weren’t attached to the spring, we can approach this by applying Newton's laws and kinematic principles.

Step 1: Net Force on the Rocket (Without Spring)

If the rocket is not attached to the spring, it only experiences the thrust force and the force of gravity. These forces affect its acceleration, which will in turn determine its velocity after traveling a specific distance.

  1. Thrust Force (FthrustF_{\text{thrust}}) = 210 N

  2. Gravitational Force (FgravityF_{\text{gravity}}) = mgmg, where:

    • m=10.9m = 10.9 kg
    • g=9.8m/s2g = 9.8 \, \text{m/s}^2

    Fgravity=10.9×9.8=106.82NF_{\text{gravity}} = 10.9 \times 9.8 = 106.82 \, \text{N}

  3. Net Force (FnetF_{\text{net}}) on the rocket:

    Fnet=FthrustFgravity=210106.82=103.18NF_{\text{net}} = F_{\text{thrust}} - F_{\text{gravity}} = 210 - 106.82 = 103.18 \, \text{N}

Step 2: Calculate Acceleration

Using Newton's second law, F=maF = ma, we can find the acceleration aa.

a=Fnetm=103.1810.99.47m/s2a = \frac{F_{\text{net}}}{m} = \frac{103.18}{10.9} \approx 9.47 \, \text{m/s}^2

Step 3: Determine the Distance Traveled by the Rocket (Based on the Spring Compression)

When the rocket is attached to the spring, the distance dd it travels is based on the compression of the spring under the thrust force. Using Hooke's law:

Fspring=kdF_{\text{spring}} = k \cdot d

Since the thrust force is balanced by the spring force at maximum compression, Fthrust=kdF_{\text{thrust}} = k \cdot d:

d=Fthrustk=2104800.4375md = \frac{F_{\text{thrust}}}{k} = \frac{210}{480} \approx 0.4375 \, \text{m}

Step 4: Use Kinematic Equation to Find Velocity After Traveling 0.4375 m

Now that we have the distance d=0.4375md = 0.4375 \, \text{m}, we can use the kinematic equation to find the final velocity vv (starting from rest):

v2=u2+2adv^2 = u^2 + 2ad

where:

  • u=0u = 0 (initial velocity),
  • a=9.47m/s2a = 9.47 \, \text{m/s}^2,
  • d=0.4375md = 0.4375 \, \text{m}.

So,

v2=0+29.470.4375v^2 = 0 + 2 \cdot 9.47 \cdot 0.4375

v28.28v^2 \approx 8.28

v8.282.88m/sv \approx \sqrt{8.28} \approx 2.88 \, \text{m/s}

Conclusion

If the rocket were not attached to the spring, its speed after traveling 0.4375 m would be approximately 2.88 m/s.

Would you like further details or have any questions?


Related Questions to Explore

  1. How would the speed change if the thrust were doubled?
  2. What effect would a higher spring constant have on the compression distance?
  3. How does gravitational force affect the rocket’s maximum height?
  4. What would be the rocket's speed after 1 meter of travel without the spring?
  5. How would adding mass to the rocket change its final speed over the same distance?

Tip

Always analyze forces separately to determine net acceleration before applying kinematic equations for accurate results.

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Math Problem Analysis

Mathematical Concepts

Newton's Laws of Motion
Kinematics
Force Analysis
Work-Energy Principle

Formulas

Net Force (F_net) = Thrust Force - Gravitational Force
F_net = ma
Hooke's Law: F_spring = k * d
Kinematic Equation: v^2 = u^2 + 2 * a * d

Theorems

Newton's Second Law of Motion
Hooke's Law

Suitable Grade Level

Grades 11-12