Math Problem Statement

a spaceship traveling through space accelerates at 0.5 m/s^2 due to force of the engine propellant, F^{} Fengine of ship \text{F}^{\text{engine of ship }}. A passing meteor shower pushes against the spaceship with a force F^{\text{ meteor on ship}}^{\text{ }}of 500 N (newtons). If the spaceship's mass is 1300 kg what is the force the engine exerts on the ship?

Solution

To solve this problem, we need to determine the net force acting on the spaceship using Newton's Second Law of Motion and then calculate the force exerted by the spaceship's engine.

The formula for net force is:

Fnet=maF_{\text{net}} = ma

Where:

  • mm is the mass of the spaceship (1300 kg),
  • aa is the acceleration of the spaceship (0.5 m/s2^2).

The net force required to produce the acceleration is:

Fnet=1300×0.5=650N.F_{\text{net}} = 1300 \times 0.5 = 650 \, \text{N}.

Now, the engine force (FengineF_{\text{engine}}) must overcome both the net force and the opposing force from the meteor shower (FmeteorF_{\text{meteor}}):

Fengine=Fnet+Fmeteor.F_{\text{engine}} = F_{\text{net}} + F_{\text{meteor}}.

Substituting the values: Fengine=650+500=1150N.F_{\text{engine}} = 650 + 500 = 1150 \, \text{N}.

Final Answer:

The engine exerts a force of 1150 N on the spaceship.

Further Questions for Exploration:

  1. What would happen to the required engine force if the mass of the spaceship were doubled?
  2. How would the acceleration change if the meteor shower exerted a larger opposing force?
  3. If the engine produced only 500 N of force, what would the resulting acceleration be?
  4. How long would it take for the spaceship to reach a velocity of 100 m/s under these conditions?
  5. Can you calculate the work done by the engine force after traveling a distance of 200 meters?

Tip: Always account for opposing forces when calculating net force in real-world scenarios!

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Math Problem Analysis

Mathematical Concepts

Newton's Second Law
Force Analysis

Formulas

F_net = m * a
F_engine = F_net + F_meteor

Theorems

Newton's Second Law of Motion

Suitable Grade Level

Grades 9-12