/*! 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 22 Draw a free-body diagram of a bl... [FREE SOLUTION] | 91影视

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Chapter 5: Problem 22

Draw a free-body diagram of a block which slides down a frictionless plane having an inclination of \(\theta=15.0^{\circ}\) (Fig. P5.22). The block starts from rest at the top and the length of the incline is \(2.00 \mathrm{m} .\) Find (a) the acceleration of the block and (b) its speed when it reaches the bottom of the incline. the block and (b) its speed when it reaches the bottom of the incline.

Short Answer

Expert verified
The acceleration of the block is: \(g \sin(\theta)\) and the speed when it reaches the bottom of the incline can be found using the kinematic equation \(v^2 = u^2 + 2as\).

Step by step solution

01

Identify and analyze the forces

The forces acting on the block are the gravitational force (\(mg\)), which can be broken into two components. One component along the incline (\(mg \sin(\theta)\)) and the other perpendicular to the incline (\(mg \cos(\theta)\)). Since the surface is frictionless, there is no frictional force acting on the block. Also there's no external force being applied.
02

Apply Newton's Second Law along the incline

According to Newton's second law, the sum of forces is equal to the mass times acceleration: \(\Sigma F = ma\). Since motion is along the direction of incline, we consider the forces along the incline. Here, the force causing the block to slide down is \(mg \sin(\theta)\), so \(mg \sin(\theta) = ma\). Simplifying we find that the acceleration \(a = g \sin(\theta)\). For \(g = 9.8 \, m/s^2\) and \(\theta = 15\degree\), it can be calculated.
03

Find the final speed using kinematic equation

With the acceleration determined in the previous step, we can use the kinematic equation \(v^2 = u^2 + 2as\) to find the final speed at the bottom of the incline. Here, initial speed (\(u\)) is zero (as the block starts from rest), acceleration (\(a\)) is the value we calculated before, and \(s\) is the distance traveled, i.e., length of the incline (2.00 m). Substituting these values, we can find the final speed.

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Key Concepts

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

Free-Body Diagram
A free-body diagram is a simple illustration used to show all the forces acting on an object. It helps in analyzing the physical situation clearly and organizing information effectively. For a block sliding down a frictionless incline, we need to consider:
  • The gravitational force, usually represented by an arrow pointing downward.
  • This force is often split into components, parallel and perpendicular to the incline:
    • Parallel component: This is responsible for the block's movement, calculated as \(mg \sin(\theta)\).
    • Perpendicular component: This acts perpendicular to the incline, written as \(mg \cos(\theta)\), balanced by the normal force.
Drawing a free-body diagram helps visualize these forces, making it easier to understand the resulting motion of the block.
Acceleration on Inclined Planes
Understanding acceleration on inclined planes involves applying Newton鈥檚 laws, specifically Newton's second law of motion. When an object slides down a frictionless incline, we consider only the force causing this movement:
  • Force along the incline: Given by \(mg \sin(\theta)\), where \(m\) is mass and \(g\) is gravitational acceleration (\(9.8 \, \text{m/s}^2\)).
  • By setting \(mg \sin(\theta) = ma\) and solving for \(a\), we find \(a = g \sin(\theta)\).
This calculation shows that an object on a frictionless incline does not depend on its mass, only on the angle of inclination \(\theta\) and gravity.
Kinematic Equations
Kinematic equations describe motion under constant acceleration. They are crucial for understanding how objects move along inclined planes. To find the final speed of a block sliding down the incline:
  • Use the equation: \(v^2 = u^2 + 2as\).
  • For a block starting from rest, \(u = 0\), making the equation \(v^2 = 2as\).
  • Substitute the known values: \(a\) from the previous section and \(s = 2.00 \, \text{m}\) as the distance.
This method provides a systematic way of predicting how fast the block will be moving at the incline's end, enhancing the understanding of motion on inclined planes.

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Most popular questions from this chapter

A Chevrolet Corvette convertible can brake to a stop from E speed of \(60.0 \mathrm{mi} / \mathrm{h}\) in a distance of \(123 \mathrm{ft}\) on a level roadway. What is its stopping distance on a roadway sloping downward at an angle of \(10.0^{\circ} ?\)

A block weighing \(75.0 \mathrm{N}\) rests on a plane inclined at \(25.0^{\circ}\) to the horizontal. A force \(F\) is applied to the object at \(40.0^{\circ}\) to the horizontal, pushing it upward on the plane. The coefficients of static and kinetic friction between the block and the plane are, respectively, 0.363 and \(0.156 .\) (a) What is the minimum value of \(F\) that will prevent the block from slipping down the plane? (b) What is the minimum value of \(F\) that will start the block moving up the plane? (c) What value of \(F\) will move the block up the plane with constant velocity?

The distance between two telephone poles is \(50.0 \mathrm{m}\) When a \(1.00-\mathrm{kg}\) bird lands on the telephone wire midway between the poles, the wire sags \(0.200 \mathrm{m}\). Draw a free-body diagram of the bird. How much tension does the bird produce in the wire? Ignore the weight of the wire.

To model a spacecraft, a toy rocket engine is securely fastened to a large puck, which can glide with negligible friction over a horizontal surface, taken as the \(x y\) plane. The \(4.00-\mathrm{~kg}\) puck has a velocity of \(300 \hat{\mathrm{i}} \mathrm{m} / \mathrm{s}\) at one instant. Eight seconds later, its velocity is to be \((800 \hat{\mathrm{i}}+10.0 \hat{\mathrm{j}}) \mathrm{m} / \mathrm{s}\) Assuming the rocket engine exerts a constant horizontal force, find (a) the components of the force and (b) its magnitude.

\- Three forces acting on an object are given by \(\mathbf{F}_{1}=(-2.00 \hat{\mathbf{i}}+2.00 \hat{\mathbf{j}}) \mathrm{N}, \quad \mathbf{F}_{2}=(5.00 \hat{\mathbf{i}}-3.00 \hat{\mathbf{j}}) \mathrm{N}, \quad\) and \(\mathbf{F}_{3}=(-45.0 \mathbf{i}) \mathrm{N} .\) The object experiences an acceleration of magnitude \(3.75 \mathrm{m} / \mathrm{s}^{2} .\) (a) What is the direction of the acceleration? (b) What is the mass of the object? (c) If the object is initially at rest, what is its speed after 10.0 s? (d) What are the velocity components of the object after \(10.0 \mathrm{s} ?\)
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