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Chapter 4: Problem 43

A skateboarder, with an initial speed of \(2.0 \mathrm{~m} / \mathrm{s}\), rolls virtually friction free down a straight incline of length \(18 \mathrm{~m}\) in \(3.3 \mathrm{~s}\). At what angle \(\theta\) is the incline oriented above the horizontal?

Short Answer

Expert verified
The incline is oriented at approximately \(7.78^\circ\).

Step by step solution

01

Calculate final velocity

To determine the final velocity, use the formula for constant acceleration: \[ v = u + at \] where \( v \) is the final velocity, \( u = 2.0 \, \mathrm{m/s} \) is the initial velocity, \( a \) is the acceleration, and \( t = 3.3 \, \mathrm{s} \) is the time. We will first determine \( a \) using the kinematic equations.
02

Determine acceleration

Using the kinematic equation:\[ s = ut + \frac{1}{2}at^2 \]where \( s = 18 \, \mathrm{m} \), \( u = 2.0 \, \mathrm{m/s} \), and \( t = 3.3 \, \mathrm{s} \), we can solve for \( a \):\[ 18 = 2.0(3.3) + \frac{1}{2}a(3.3)^2 \]. Solve for \( a \) which results in \( a \approx 1.33 \, \mathrm{m/s^2} \).
03

Calculate final speed using the acceleration

Substitute the acceleration \( a = 1.33 \, \mathrm{m/s^2} \) back into the velocity equation:\[ v = 2.0 + (1.33)(3.3) = 6.39 \, \mathrm{m/s} \].This gives us the final speed \( v \).
04

Determine the angle of the incline

Using the equation for motion along an incline \[ a = g \sin(\theta) \], where \( g = 9.81 \, \mathrm{m/s^2} \), solve for \( \theta \):\[ 1.33 = 9.81 \sin(\theta) \]\[ \sin(\theta) = \frac{1.33}{9.81} \]\[ \theta = \arcsin\left(\frac{1.33}{9.81}\right) \approx 7.78^\circ \].So, the angle \( \theta \) is approximately \( 7.78^\circ \).

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Key Concepts

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

Incline Motion
Incline motion refers to the movement of an object along a sloped surface. Unlike horizontal motion, incline motion is influenced by the force of gravity acting down the slope. This causes the object to accelerate or decelerate depending on the direction of movement. When understanding incline motion, it's crucial to consider:
  • The force components along and perpendicular to the incline. Gravity acts downwards, creating a component that pulls the object along the slope.
  • The "normal force," which acts perpendicular to the surface of the incline, playing a role in determining friction, which can often be ignored if stated, such as in 'virtually friction-free' scenarios like this one.
  • With no friction, the motion is solely dictated by gravity, simplifying analysis. This lets the object move freely down the slope.
By analyzing these factors, we can predict how fast an object will move along the incline under different initial conditions, such as speed or height.
Kinematic Equations
Kinematic equations are a set of formulas used to describe the motion of objects under constant acceleration. They are particularly useful in analyzing incline motion, as they provide a systematic way to calculate variables such as velocity, acceleration, or distance.
  • The most commonly used kinematic equations include:
    • \( v = u + at \)- This describes the final velocity \( v \) as a function of initial velocity \( u \), acceleration \( a \), and time \( t \).
    • \( s = ut + \frac{1}{2}at^2 \)- This calculates displacement \( s \), incorporating initial velocity, acceleration, and the square of time.
    • \( v^2 = u^2 + 2as \)- This relates the square of velocities to acceleration and displacement.
In our given scenario, these equations help calculate the skateboarder's acceleration and final velocity after descending the incline. By using the initial velocity, time spent on the incline, and the total distance, we precisely determine motion characteristics.
Incline Angle Calculation
Once the acceleration of the skateboarder has been determined, the next step is to find the angle of the incline, often denoted as \( \theta \). This angle affects how gravity accelerates the object down the slope.
  • The relationship for acceleration along an incline is expressed as:
    • \( a = g \sin(\theta) \)
  • Here, \( g \) is the acceleration due to gravity \( 9.81 \, \mathrm{m/s^2} \), and \( \sin(\theta) \) gives the ratio of the incline's height over its length.
To solve for \( \theta \):
  • Rearrange the formula to find \( \sin(\theta) = \frac{a}{g} \).
  • Use the inverse sine function (\( \arcsin \)) to calculate the angle: \( \theta = \arcsin\left(\frac{a}{g}\right) \).
  • In the exercise, substituting the values gives \( \theta \approx 7.78^\circ \).
This calculation is crucial as it tells us how steep the incline must be for the given motion, linking the physics of angles with observable motion.

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Most popular questions from this chapter

A super high-speed 14-car Italian train has a mass of 640 metric tons \((640,000 \mathrm{~kg})\). It can exert a maximum force of \(400 \mathrm{kN}\) horizontally against the tracks, whereas at maximum constant velocity \((300 \mathrm{~km} / \mathrm{h})\), it exerts a force of about \(150 \mathrm{kN}\). Calculate \((a)\) its maximum acceleration, and \((b)\) estimate the force of friction and air resistance at top speed.

A particular race car can cover a quarter-mile track \((402 \mathrm{~m})\) in \(6.40 \mathrm{~s}\) starting from a standstill. Assuming the acceleration is constant, how many "g's" does the driver experience? If the combined mass of the driver and race car is \(535 \mathrm{~kg},\) what horizontal force must the road exert on the tires?

A person has a reasonable chance of surviving an automobile crash if the deceleration is no more than 30 g's. Calculate the force on a \(65-\mathrm{kg}\) person accelerating at this rate. What distance is traveled if brought to rest at this rate from \(95 \mathrm{~km} / \mathrm{h} ?\)

A 7650 -kg helicopter accelerates upward at \(0.80 \mathrm{~m} / \mathrm{s}^{2}\) while lifting a \(1250-\mathrm{kg}\) frame at a construction site, Fig. \(4-59 .\) (a) What is the lift force exerted by the air on the helicopter rotors? (b) What is the tension in the cable (ignore its mass) that connects the frame to the helicopter? (c) What force does the cable exert on the helicopter?

How much tension must a rope withstand if it is used to accelerate a \(1210-\mathrm{kg}\) car horizontally along a frictionless surface at \(1.20 \mathrm{~m} / \mathrm{s}^{2} ?\)
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