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Chapter 5: Problem 12

A 125 -kg (including all the contents) rocket has an engine that produces a constant vertical force (the thrust) of 1720 \(\mathrm{N}\) . Inside this rocket, a 15.5 -N electrical power supply rests on the floor. (a) Find the acceleration of the rocket. (b) When it has reached an altitude of \(120 \mathrm{m},\) how hard does the floor push on the power supply? (Hint: Start with a free-body diagram for the power supply.)

Short Answer

Expert verified
(a) Rocket's acceleration: 3.96 m/s虏. (b) Floor's force on power supply: 21.75 N.

Step by step solution

01

Calculate Net Force

First, identify the forces acting on the rocket. The thrust force is 1720 N upward, and the gravitational force is acting downward. Calculate the gravitational force using the formula \( F_{gravity} = m imes g \), where \( m = 125 \) kg (mass of the rocket) and \( g = 9.8 \) m/s虏 (acceleration due to gravity). \[ F_{gravity} = 125 \times 9.8 = 1225 \text{ N} \] The net force is the thrust force minus the gravitational force: \( F_{net} = 1720 - 1225 = 495 \text{ N} \).
02

Calculate Rocket's Acceleration

Use Newton鈥檚 second law to find the rocket's acceleration: \( F_{net} = m imes a \).Rearrange the formula to solve for acceleration \( a \): \( a = \frac{F_{net}}{m} \). Using \( F_{net} = 495 \text{ N} \) and \( m = 125 \text{ kg} \): \[ a = \frac{495}{125} = 3.96 \text{ m/s}^2 \].
03

Analyze Free-Body Diagram for the Power Supply

Consider the forces acting on the 15.5 N electrical power supply. The forces include the normal force \( N \) (floor's push on the power supply) and gravitational force \( F_g = 15.5 \text{ N} \) (downward). The power supply is inside the rocket, moving with acceleration \( a = 3.96 \text{ m/s}^2 \).
04

Apply Newton鈥檚 Second Law on Power Supply

For the power supply: \( N - F_g = m_{ps} \times a \), where \( m_{ps} \) is the mass of the power supply.First, calculate \( m_{ps} \) using \( F_g = m_{ps} imes g \): \( m_{ps} = \frac{15.5}{9.8} \approx 1.58 \text{ kg} \).Substitute \( m_{ps} \) and \( a \) into the equation: \[ N - 15.5 = 1.58 \times 3.96 \] \[ N = 6.253 + 15.5 \] \[ N = 21.753 \text{ N} \].
05

Conclusion

The acceleration of the rocket is \( 3.96 \text{ m/s}^2 \). At an altitude of 120 m, the floor of the rocket exerts a force of about \( 21.75 \text{ N} \) on the power supply.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Rocket Propulsion
Rocket propulsion is an intriguing concept that applies Newton's Second Law of Motion, which states that the acceleration of an object depends on the net force acting upon it and the object's mass. For a rocket, the engine produces a thrust force that propels it vertically upward.
As the rocket travels upward, this thrust has to overcome the gravitational force that pulls it downward to achieve lift-off. This is mathematically represented as:
  • Thrust force (upward) and gravitational force (downward) oppose each other.
  • The net force, which determines the rocket's movement, is the difference between these forces.
The thrust force accelerates the rocket, allowing it to ascend into the sky by producing a force greater than the gravitational pull. This net upward force results in the acceleration calculated from the formula, a central component of rocket science theory.
Understanding thrust and rocket propulsion involves realizing how engines are engineered to exert greater force to compensate for the rocket鈥檚 mass and the constant force of gravity.
Free-Body Diagram
A free-body diagram is a useful tool in physics to visualize the different forces acting on an object. In this case, the diagram helps us understand how the forces interact on the power supply inside the rocket.
To draw a free-body diagram for the power supply, pay attention to:
  • The downward gravitational force, which in this scenario is 15.5 N.
  • The normal force, which is the force exerted by the floor of the rocket on the power supply to support it in the upward acceleration.
The power supply isn't accelerating independently; instead, it moves with the same acceleration as the rocket, which we calculated to be 3.96 m/s虏.
This method allows us to determine how much force the floor of the rocket exerts on the power supply, termed the normal force, which is crucial for understanding interaction forces within accelerating systems.
Gravity Force Calculation
Calculating the force due to gravity is essential when analyzing any movement, especially in a vertical motion like that of a rocket. The gravitational force is how much the Earth pulls on an object and is measured using the equation:
  • \( F_{gravity} = m \times g \)
  • where \( m \) is the mass of the object and \( g \) is the acceleration due to gravity.
In our rocket exercise, we considered two gravitational force calculations:
  • One for the rocket, weighing 125 kg, which calculated to be 1225 N.
  • Another for the electrical power supply inside, weighing about 1.58 kg, resulting in a 15.5 N gravitational pull.
These calculations are significant as they allow us to distinguish the forces that need to be overcome for the rocket to climb upwards. Understanding gravity's role helps us accurately compute other forces and anticipate movement dynamics in physics and engineering scenarios.

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Most popular questions from this chapter

A 3.00 -kg box that is several hundred meters above the surface of the earth is suspended from the end of a short vertical rope of negligible mass. A time-dependent upward force is applied to the upper end of the rope, and this results in a tension in the rope of \(T(t)=(36.0 \mathrm{N} / \mathrm{s}) t .\) The box is at rest at \(t=0 .\) The only forces on the box are the tension in the rope and gravity. (a) What is the velocity of the box at ( i \(t=1.00 \mathrm{s}\) and (ii) \(t=3.00 \mathrm{s} ?\) (b) What is the maximum distance that the box descends below its initial position? (c) At what value of \(t\) does the box return to its initial position?

Block \(A\) in Fig. P5.72 weighs 60.0 \(\mathrm{N} .\) The coefficient of static friction between the block and the surface on which it rests is \(0.25 .\) The weight \(w\) is 12.0 \(\mathrm{N}\) and the system is in equilibrium. (a) Find the friction force exerted on block \(A .\) (b) Find the maximum weight \(w\) for which the system will remain in equilibrium.

A physics major is working to pay his college tuition by performing in a traveling carnival. He rides a motorcycle inside a hollow, transparent plastic sphere. After gaining sufficient speed, he travels in a vertical circle with a radius of 13.0 \(\mathrm{m} .\) The physics major has mass \(70.0 \mathrm{kg},\) and his motorcycle has mass 40.0 \(\mathrm{kg}\) . (a) What minimum speed must he have at the top of the circle if the tires of the motorcycle are not to lose contact with the sphere? (b) At the bottom of the circle, his speed is twice the value calculated in part (a). What is the magnitude of the normal force exerted on the motorcycle by the sphere at this point?

A small remote-controlled car with mass 1.60 \(\mathrm{kg}\) moves at a constant speed of \(v=12.0 \mathrm{m} / \mathrm{s}\) in a vertical circle inside a hollow metal cylinder that has a radius of 5.00 \(\mathrm{m}(\) Fig. \(\mathrm{P} 5.120) .\) What is the magnitude of the normal force exerted on the car by the walls of the cylinder at (a) point \(A\) (at the bottom of the vertical circle) and (b) point \(B(\) at the top of the vertical circle)?

People who do chinups raise their chin just over a bar (the chinning bar), supporting themselves with only their arms. Typically, the body below the arms is raised by about 30 \(\mathrm{cm}\) in a time of 1.0s, starting from rest. Assume that the entire body of a \(680-\mathrm{N}\) person doing chin-ups is raised this distance and that half the 1.0 \(\mathrm{s}\) is spent accelerating upward and the other half accelerating downward, uniformly in both cases. Draw a free-body diagram of the person's body, and then apply it to find the force his arms must exert on him during the accelerating part of the chin-up.
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