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A crate of oranges slides down an inclined plane without friction. If it is released from rest and reaches a speed of \(5.832 \mathrm{~m} / \mathrm{s}\) after sliding a distance of \(2.29 \mathrm{~m}\), what is the angle of inclination of the plane with respect to the horizontal?

Short Answer

Expert verified
Answer: The angle of inclination of the plane with respect to the horizontal is 47.4°.

Step by step solution

01

We will use the following equation of motion to relate the initial and final speeds, the distance traveled, and the acceleration: \(v^2 = u^2 + 2as\) where \(v\) is the final speed, \(u\) is the initial speed, \(a\) is the acceleration, and \(s\) is the distance traveled. Additionally, we will use trigonometry to find the angle of inclination, using the acceleration component along the incline. We are given: Initial speed \(u = 0 \mathrm{~m} / \mathrm{s}\) (released from rest) Final speed \(v = 5.832 \mathrm{~m} / \mathrm{s}\) Distance traveled \(s = 2.29 \mathrm{~m}\) We need to find: Angle of inclination \(\theta\) #Step 2: Calculate the acceleration along the incline#

First, we will use the equation of motion: \(v^2 = u^2 + 2as\) Plugging in the values, we have: \((5.832)^2 = 0^2 + 2a(2.29)\) Now, solve for \(a\): \(a = \frac{(5.832)^2}{2(2.29)}\) Calculating the value of \(a\), we get: \(a = 7.176 \mathrm{~m}/\mathrm{s}^2\) #Step 3: Relate the acceleration to the angle of inclination#
02

The acceleration along the incline is related to the gravitational acceleration (\(g = 9.81 \mathrm{~m}/\mathrm{s}^2\)) and the angle of inclination \(\theta\) by the following equation: \(a = g \cdot \sin(\theta)\) Now, plug in the values of \(a\) and \(g\) into the equation: \(7.176 = 9.81 \cdot \sin(\theta)\) Now, solve for \(\sin(\theta)\): \(\sin(\theta) = \frac{7.176}{9.81}\) Calculating the value of \(\sin(\theta)\), we get: \(\sin(\theta) = 0.732\) #Step 4: Calculate the angle of inclination#

To find the angle of inclination \(\theta\), take the inverse sin of the value we found for \(\sin(\theta)\): \(\theta = \arcsin(0.732)\) Calculating the angle, we get: \(\theta = 47.4^\circ\) So the angle of inclination of the plane with respect to the horizontal is \(47.4^\circ\).

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Key Concepts

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

Inclined Plane Physics
Understanding the physics of an inclined plane can be incredibly intriguing as it combines principles of mechanics and trigonometry. Simply put, an inclined plane is a flat surface tilted at an angle to the horizontal. This arrangement creates a unique situation where the force of gravity acts in two distinct components – one perpendicular to the plane and the other parallel to it.

The parallel component is responsible for the acceleration of an object down the slope when there's no friction. While calculating such motion, we assume frictionless conditions to simplify our equations, although in real life, friction could significantly affect the outcome. In other words, we can predict an object’s velocity after it slides down a known distance if we know the incline's angle.

In the exercise provided, by knowing the final speed and the distance traveled by the crate of oranges, we could determine the gravitational acceleration component along the slope. Students can improve their understanding by visualizing the inclined plane and breaking down the forces, which could help assimilate the core concepts intuitively.
Kinematics Equations
Kinematics is the branch of physics that deals with motion without considering the forces causing the motion. A fundamental aspect of kinematics in classical mechanics is the use of kinematic equations to describe the motion of an object. These equations relate displacement, initial velocity, final velocity, acceleration, and time.

One of these equations is the one used in the exercise: \(v^2 = u^2 + 2as\). This particular equation enables us to calculate an object's final velocity (\(v\)) without knowing the time it took to reach that velocity, which is often the case with problems involving inclined planes. To fully grasp this equation's application, it is important to perform a series of practice problems and understand its derivation, which stems from integrating acceleration respect to time.

When applying these equations, ensure that all the units are consistent and understand that acceleration is constant, which is a common trait in kinematic problems.
Trigonometry in Physics
Trigonometry plays a pivotal role in physics, especially when it comes to analyzing forces, motion, and angles. Most students find the interplay between trigonometry and physics in scenarios like the inclined plane. To resolve the motion into components or find the angle of inclination, trigonometric functions such as sine, cosine, and tangent are employed.

In our exercise, the angle of inclination is found using the sine function, which in a right-angled triangle, relates the ratio of the length of the opposite side to the hypotenuse. This concept was crucial to relate the gravitational force parallel to the inclined plane to the acceleration of the crate. Students can benefit by deeply understanding how these functions are derived from the properties of a right triangle, and frequently applying them to different physics problems involving angles and motion.

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Most popular questions from this chapter

A bowling ball of mass \(M_{1}=6.00 \mathrm{~kg}\) is initially at rest on the sloped side of a wedge of mass \(M_{2}=9.00 \mathrm{~kg}\) that is on a frictionless horizontal floor. The side of the wedge is sloped at an angle of \(\theta=36.9^{\circ}\) above the horizontal. a) What is the magnitude of the horizontal force that should be exerted on the bowling ball to keep it at a constant height on the slope? b) What is the magnitude of the acceleration of the wedge, if no external force is applied?

A load of bricks of mass \(M=200.0 \mathrm{~kg}\) is attached to a crane by a cable of negligible mass and length \(L=3.00 \mathrm{~m}\). Initially, when the cable hangs vertically downward, the bricks are a horizontal distance \(D=1.50 \mathrm{~m}\) from the wall where the bricks are to be placed. What is the magnitude of the horizontal force that must be applied to the load of bricks (without moving the crane) so that the bricks will rest directly above the wall?

4.27 The gravitational acceleration on the Moon is a sixth of that on Earth. The weight of an apple is \(1.00 \mathrm{~N}\) on Earth. a) What is the weight of the apple on the Moon? b) What is the mass of the apple?

A piñata of mass \(M=12.0 \mathrm{~kg}\) hangs on a rope of negligible mass that is strung between the tops of two vertical poles. The horizontal distance between the poles is \(D=2.00 \mathrm{~m},\) the top of the right pole is a vertical distance \(h=0.500 \mathrm{~m}\) higher than the top of the left pole, and the total length of the rope between the poles is \(L=3.00 \mathrm{~m} .\) The piñata is attached to a ring, with the rope passing through the center of the ring. The ring is frictionless, so that it can slide freely along the rope until the piñata comes to a point of static equilibrium. a) Determine the distance from the top of the left (lower) pole to the ring when the pinata is in static equilibrium. b) What is the tension in the rope when the pinata is at this point of static equilibrium?

A tractor pulls a sled of mass \(M=1000\). kg across level ground. The coefficient of kinetic friction between the sled and the ground is \(\mu_{\mathrm{k}}=0.600 .\) The tractor pulls the sled by a rope that connects to the sled at an angle of \(\theta=30.0^{\circ}\) above the horizontal. What magnitude of tension in the rope is necessary to move the sled horizontally with an acceleration \(a=2.00 \mathrm{~m} / \mathrm{s}^{2}\) ?

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