/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 58 A block is projected up a fricti... [FREE SOLUTION] | 91影视

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Chapter 6: Problem 58

A block is projected up a frictionless inclined plane with initial speed \(v_{1}=3.50 \mathrm{~m} / \mathrm{s}\). The angle of incline is \(\theta=32.0^{\circ}\). (a) How far up the plane does it go? (b) How long does it take to get there? (c) What is its speed when it gets back to the bottom?

Short Answer

Expert verified
(a) 1.18 m (b) 0.674 s (c) 3.50 m/s

Step by step solution

01

- Determine acceleration due to gravity along the incline

Find the acceleration component along the incline using the formula: \( a = g \sin(\theta) \) where \( g = 9.8 \ m/s^2 \ \ and \ \ \theta = 32.0^{\circ} \). Substitute to get: \[ a = 9.8 \sin(32.0^{\circ}) \approx 5.19 \ \mathrm{m/s^2} \].
02

- Calculate the distance up the plane

Use the kinematic equation: \[ v_f^2 = v_1^2 - 2ad \] to find distance \(d\), where \(v_f = 0 \) (since the block comes to a stop), \( v_1 = 3.50 \ \mathrm{m/s} \) and \(a = 5.19 \ \mathrm{m/s^2} \). Rearrange the equation to get: \[ d = \frac{{v_1^2}}{{2a}} = \frac{{(3.50)^2}}{{2 \times 5.19}} \approx 1.18 \ \mathrm{m} \].
03

- Calculate the time to reach the highest point

Use another kinematic equation: \[ v_f = v_1 - at \ \Rightarrow t = \frac{{v_1}}{{a}} \]. Substitute the values to get: \[ t = \frac{{3.50}}{{5.19}} \approx 0.674 \ \mathrm{s} \].
04

- Determine speed at the bottom

Since the plane is frictionless, the block's speed will be the same when it returns to the bottom due to conservation of energy. Thus, the speed is: \[ v = 3.50 \mathrm{\ m/s} \].

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Key Concepts

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

Inclined Plane Physics
An inclined plane is a flat surface tilted at an angle, different from the horizontal. It is a simple yet powerful concept in physics, often used to understand motion and forces. When an object slides on this tilted surface, gravity causes it to accelerate down the slope. To find the acceleration along the incline, we can use the component of gravitational force parallel to the plane. This component is calculated using the formula: \( a = g \sin(\theta) \), where \( g \approx 9.8 \mathrm{m/s^2} \) is the acceleration due to gravity, and \( \theta \) is the angle of the incline. In the provided exercise, the angle \( 32.0^{\circ} \) leads to an acceleration of approximately \( 5.19 \mathrm{m/s^2} \) down the plane. This value assists in determining further motion details, such as the distance traveled and time taken.
Acceleration Due to Gravity
Gravity is a fundamental force acting on all objects with mass. On Earth, it causes objects to accelerate downward at a rate of \( 9.8 \mathrm{m/s^2} \). When an object moves along an inclined plane, only a portion of this gravitational force acts to accelerate the object down the slope. This portion is determined by the sine of the incline's angle, symbolized as \( g \sin(\theta) \). For the block in the exercise, this results in an acceleration of \( 5.19 \mathrm{m/s^2} \) down the inclined plane. Understanding this is essential for solving various physics problems related to motion on slopes, calculating the distance traveled or the time taken can be simplified using kinematic equations in these scenarios. This principle allows us to analyze the motion of objects on slopes accurately.
Conservation of Energy
The principle of conservation of energy states that energy in a closed system remains constant. Energy can neither be created nor destroyed, but it can be transformed from one form to another. In the context of the provided exercise, the block's kinetic energy at the beginning (\( \frac{{1}}{{2}}mv_1^2 \)) is first converted to potential energy (\( mgh \)) as it moves up the plane and then back to kinetic energy as it returns to its original position. As the plane is frictionless, there are no energy losses to friction, meaning the block's speed when it returns to the bottom must be the same as its initial projection speed, \( 3.50 \mathrm{m/s} \). This shows the practical application of energy conservation in analyzing the block's motion without needing to track every vector and force involved in the process.

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Most popular questions from this chapter

A \(3.5 \mathrm{~kg}\) block is pushed along a horizontal floor by a force \(\vec{F}\) of magnitude \(15 \mathrm{~N}\) at an angle \(\theta=40^{\circ}\) with the horizontal (Fig. 6-49). The coefficient of kinetic friction between the block and the floor is \(0.25 .\) Calculate the magnitudes of (a) the frictional force on the block from the floor and (b) the acceleration of the block.

A \(1000 \mathrm{~kg}\) boat is traveling at \(90 \mathrm{~km} / \mathrm{h}\) when its engine is shut off. The magnitude of the frictional force \(\vec{f}^{\text {kin }}\) between boat and water is proportional to the speed \(v\) of the boat; \(\vec{f}^{\mathrm{kin}}=(70 \mathrm{~N} \cdot \mathrm{s} / \mathrm{m}) \vec{v}\). Find the time required for the boat to slow to \(45 \mathrm{~km} / \mathrm{h}\).

For sport, a \(12 \mathrm{~kg}\) armadillo runs onto a large pond of level, frictionless ice with an initial velocity of \(5.0 \mathrm{~m} / \mathrm{s}\) along the positive direction of an \(x\) axis. Take its initial position on the ice as being the origin. It slips over the ice while being pushed by a wind with a force of \(17 \mathrm{~N}\) in the positive direction of the \(y\) axis. In unit-vector notation, what are the animal's (a) velocity and (b) position vector when it has slid for \(3.0 \mathrm{~s}\) ?

A worker wishes to pile a cone of sand onto a circular area in his yard. The radius of the circle is \(R\), and no sand is to spill onto the surrounding area (Fig. 6-46). If \(\mu^{\text {stat }}\) is the static coefficient of friction between each layer of sand along the slope and the sand beneath it (along which is might slip), show that the greatest volume of sand that can be stored in this manner is \(\pi \mu^{\text {stat }} R^{3} / 3\). (The volume of a cone is \(A h / 3\), where \(A\) is the base area and \(h\) is the cone's height.)

A motorcycle and \(60.0 \mathrm{~kg}\) rider accelerate at \(3.0 \mathrm{~m} / \mathrm{s}^{2}\) up a ramp inclined \(10^{\circ}\) above the horizontal. (a) What is the magnitude of the net force acting on the rider? (b) What is the magnitude of the force on the rider from the motorcycle?
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