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\(\mathrm{A} 1500 \mathrm{~kg}\) rocket is to be launched with an initial upward speed of \(50.0 \mathrm{~m} / \mathrm{s} .\) In order to assist its engines, the engineers will start it from rest on a ramp that rises \(53^{\circ}\) above the horizontal (Fig. \(\mathbf{P 7 . 5 0}\) ). At the bottom, the ramp turns upward and launches the rocket vertically. The engines provide a constant forward thrust of \(2000 \mathrm{~N}\), and friction with the ramp surface is a constant \(500 \mathrm{~N}\). How far from the base of the ramp should the \text { rocket start, as measured along the surface of the ramp?

Short Answer

Expert verified
The rocket should start a distance, \(s\) from the base of the ramp, which can be calculated from the rearranged equation of motion. Fill in numbers to get the value of \(s\).

Step by step solution

01

Identify Forces

Firstly, identify all the forces acting on the rocket. These would include the force of gravity acting vertically downward, the thrust generated by the engines acting along the direction of motion and friction acting against the direction of motion. The force of gravity can be broken down into its components perpendicular and parallel to the incline.
02

Resolve Forces

Now, resolve all the forces acting on the rocket in the direction of motion (along the slope). Net force is the sum of all forces in the direction of motion. Here the main forces are the thrust of \(2000N\) by the engine, the component of gravitational force along the ramp, which is \(mg \sin 53^\circ\), and the friction which is \(500N\). So, the net force, \(F_{net}\) can be calculated as \(F_{net} = 2000N - mg \sin 53^\circ - 500N\).
03

Apply Second Law of Motion

Use the second law of motion, which states that the change in momentum of a body is directly proportional to the force applied and occurs in the direction in which the force is applied. Mathematically, \(F_{net} = ma\), where \(m\) is the mass and \(a\) is acceleration. Using this equation, we can calculate the acceleration, \(a = \frac{F_{net}}{m}\).
04

Apply Kinematic Equations

Now, apply the third equation of motion, which states that \(v^2 = u^2 + 2as\), where \(v\) is the final velocity, \(u\) is the initial velocity, \(a\) is the acceleration and \(s\) is the distance. Rearranging for \(s\), we get \(s = \frac{v^2 - u^2}{2a}\). The initial velocity \(u\) is 0, final velocity \(v\) is 50.0 m/s and \(a\) is calculated in the previous step. Substitute all known values into the equation to find the value of \(s\), the distance the rocket should start from the base of the ramp.

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Key Concepts

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

Newton's second law
Newton's second law is a fundamental principle in physics which states that the acceleration of an object is directly proportional to the net force acting upon it and inversely proportional to its mass. Mathematically, this is expressed as \(F = ma\), where \(F\) is the net force, \(m\) is the mass of the object, and \(a\) is its acceleration. Understanding this law is crucial, particularly when dealing with rocket motion. A rocket on an inclined plane involves different forces, including thrust, friction, and gravitational forces.
  • Thrust: This is the forward force produced by the rocket's engines, pushing it upward along the slope of the ramp.
  • Gravitational Force: This acts downward, needing decomposition into components parallel and perpendicular to the surface of the ramp.
  • Friction: This acts against the movement along the ramp, causing resistance to the forward thrust.
To calculate the rocket's acceleration on the ramp, the net force acting along the incline must be determined first. By summing up all the forces in the direction along the ramp's surface, the formula \(F_{net} = 2000 \text{ N} - mg \sin 53^\circ - 500 \text{ N}\) is used. Here, \(mg \sin 53^\circ\) represents the component of the gravitational force. Calculating with this net force provides the acceleration using \(a = \frac{F_{net}}{m}\). Thus, without this law, we wouldn't be able to determine how quickly the rocket can accelerate up the ramp.
kinematic equations
Kinematic equations are essential tools in physics, especially when analyzing motion with constant acceleration. In this exercise, we focus on determining how far the rocket should travel up the ramp before launch using these equations.Let's consider the kinematic equation: \(v^2 = u^2 + 2as\). Here, \(v\) represents the final velocity the rocket achieves before vertical launch, \(u\) is the initial velocity (which is zero in this scenario since the rocket starts from rest), \(a\) is the previously calculated acceleration, and \(s\) is the distance we need to find.
  • Set the final velocity \(v = 50 \text{ m/s}\), since the rocket needs to reach this speed at launch.
  • The initial velocity \(u = 0\), as given in the problem.
  • Use the calculated acceleration from Newton’s second law step.
Rearranging the equation for \(s\) gives \(s = \frac{v^2 - u^2}{2a}\). By placing all known values into this formula, students can calculate the effective distance for the rocket's ramp motion. This equation is invaluable in scenarios where an object's velocity with constant acceleration over a distance is analyzed.
inclined plane forces
Inclined plane forces are pivotal in understanding how objects move along slopes or ramps. In this problem involving rocket motion, such forces must be dissected carefully to comprehend the dynamics involved. The inclined plane creates unique scenarios where gravitational force acting on an object is split into two components:
  • Parallel component: Acts along the incline, affecting the downward sliding tendency of the object through \(mg \sin\theta\), where \(\theta\) is the angle of the incline (\(53^\circ\) in this case).
  • Perpendicular component: Acts normal to the incline, denoted as \(mg \cos\theta\). Though it doesn't influence the motion directly, it can affect normal force calculations, particularly when considering friction.
The thrust provided by the rocket's engine assists in counteracting these forces, enabling upward motion. Additionally, frictional force, constant at \(500 \text{ N}\), opposes this movement, imitating real-world conditions where surfaces are never completely smooth.By understanding these forces, students gain insight into how real-world physics problems can be simplified using components, thus making complex systems manageable and solvable through analysis and application of classical mechanics principles.

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Most popular questions from this chapter

\(\mathrm{A} 60.0 \mathrm{~kg}\) skier starts from rest at the top of a ski slope \(65.0 \mathrm{~m}\) high. (a) If friction forces do \(-10.5 \mathrm{~kJ}\) of work on her as she descends, how fast is she going at the bottom of the slope? (b) Now moving horizontally, the skier crosses a patch of soft snow where \(\mu_{\mathrm{k}}=0.20 .\) If the patch is \(82.0 \mathrm{~m}\) wide and the average force of air resistance on the skier is \(160 \mathrm{~N}\), how fast is she going after crossing the patch? (c) The skier hits a snowdrift and penetrates \(2.5 \mathrm{~m}\) into it before coming to a stop. What is the average force exerted on her by the snowdrift as it stops her?

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