/*! 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 49 In Fig. 5-45, a block of mass \(... [FREE SOLUTION] | 91影视

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Chapter 5: Problem 49

In Fig. 5-45, a block of mass \(m=5.00 \mathrm{~kg}\) is pulled along a horizontal frictionless floor by a cord that exerts a force of magnitude \(F=12.0 \mathrm{~N}\) at an angle \(\theta=25.0^{\circ} .\) (a) What is the magnitude of the block's acceleration? (b) The force magnitude \(F\) is slowly increased. What is its value just before the block is lifted (completely) off the floor? (c) What is the magnitude of the block's acceleration just before it is lifted (completely) off the floor?

Short Answer

Expert verified
(a) 2.175 m/s虏. (b) 116.094 N. (c) 21.0574 m/s虏.

Step by step solution

01

Resolve the Force Components

The force exerted by the cord is at an angle, so we need to find its horizontal and vertical components. The horizontal component is given by \( F_x = F \cos \theta \) and the vertical component is \( F_y = F \sin \theta \). Plugging in the values, we get: \( F_x = 12.0 \cdot \cos 25.0^{\circ} \) and \( F_y = 12.0 \cdot \sin 25.0^{\circ} \).
02

Calculate Horizontal Component

Using a calculator, find \( \cos 25.0^{\circ} \approx 0.9063 \). Therefore, \( F_x = 12.0 \times 0.9063 \approx 10.8756 \) N.
03

Calculate Acceleration (a)

The horizontal component of the force determines the acceleration. Using Newton's second law, \( a = \frac{F_x}{m} \). Thus, \( a = \frac{10.8756}{5.00} \approx 2.175 \) m/s虏.
04

Determine Condition for Lifting

The block will lift off the floor when the normal force is zero. That means the vertical component of the exerted force equals the gravitational force on the block, \( F_y = mg \).
05

Solve for Force Required to Lift (b)

Here, set \( F \sin \theta = mg \). Using \( \sin 25.0^{\circ} \approx 0.4226 \), solve \( F = \frac{mg}{\sin 25.0^{\circ}} = \frac{5.00 \times 9.81}{0.4226} \approx 116.094 \) N.
06

Calculate Acceleration Before Lift (c)

With \( F = 116.094 \) N, the horizontal component is \( F_x = 116.094 \cos 25.0^{\circ} \approx 105.287 \) N. The acceleration is \( a = \frac{105.287}{5.00} = 21.0574 \) m/s虏.

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Key Concepts

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

Newton's Laws of Motion
Newton's laws of motion form the foundation for understanding how forces and movement interact. In this exercise, we primarily use Newton's second law, which ties force, mass, and acceleration together through the equation \( F = ma \). This equation tells us that if you know the net force acting on an object and its mass, you can determine the acceleration.
  • First law states that an object at rest stays at rest, and an object in motion stays in motion unless acted upon by a net external force.
  • The second law, which is central to our problem, shows how acceleration (change in motion) results from a net applied force.
  • The third law states every action has an equal and opposite reaction.
By applying these laws, students can predict and explain changes in motion based on various forces acting upon an object. Remember, understanding these principles helps solve complex problems like determining acceleration or lifting forces, as given in this exercise.
Force Resolution
In the context of the given problem, force resolution involves breaking a force vector into its horizontal and vertical components. This is crucial because different components affect the block's motion in different ways.

To resolve the force applied by the cord:
  • Horizontal Component \( F_x \): Calculated using \( F_x = F \cos \theta \). It influences the horizontal motion, dictating acceleration across the frictionless surface in this scenario.
  • Vertical Component \( F_y \): Given by \( F_y = F \sin \theta \). It acts upwards and can counteract the gravitational pull, potentially lifting the block if it matches the gravitational force \( mg \).
Breaking down the force in this way provides clarity on how different parts of a force affect motion and conditions for various physical effects, such as beginning to lift off.
Acceleration Calculation
Acceleration calculation involves determining how fast the velocity of an object changes under the influence of a net force. With Newton's second law, this calculation becomes straightforward.

When calculating acceleration, use the formula \( a = \frac{F_x}{m} \):
  • The net force acting in a particular direction (here the horizontal force component \( F_x \)) is essential for calculating acceleration.
  • In our exercise, this translates to \( a = \frac{10.8756\, \text{N}}{5.00\, \text{kg}} \approx 2.175\, \text{m/s}^2 \).
Remember, the accuracy of this step hinges on correctly identifying the force components and understanding the material鈥檚 interaction with forces applied.
Frictionless Surfaces
A frictionless surface is an idealized concept where there is no opposing force due to friction, allowing a straightforward application of Newton's laws. In this exercise, the absence of friction simplifies the analysis:
  • The force applied translates directly into acceleration without any loss due to frictional forces.
  • Understanding real vs. ideal scenarios helps appreciate how friction normally retards motion, affecting the net force and, thus, the acceleration of objects.
This idealization allows students to focus on the core physics principles, making it easier to comprehend how different forces affect acceleration and motion. In reality, friction often plays a significant role, so recognizing when it can be ignored is key in solving applicable physics problems.

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Most popular questions from this chapter

The Zacchini family was renowned for their human-cannonball act in which a family member was shot from a cannon using either elastic bands or compressed air. In one version of the act, Emanuel Zacchini was shot over three Ferris wheels to land in a net at the same height as the open end of the cannon and at a range of \(69 \mathrm{~m} .\) He was propelled inside the barrel for \(5.2 \mathrm{~m}\) and launched at an angle of \(53^{\circ} .\) If his mass was \(85 \mathrm{~kg}\) and he underwent constant acceleration inside the barrel, what was the magnitude of the force propelling him? (Hint: Treat the launch as though it were along a ramp at \(53^{\circ} .\) Neglect air drag.)

If the 1 kg standard body is accelerated by only \(\vec{F}_{1}=(3.0 \mathrm{~N}) \hat{\mathrm{i}}+(4.0 \mathrm{~N}) \hat{\mathrm{j}}\) and \(\vec{F}_{2}=(-2.0 \mathrm{~N}) \hat{\mathrm{i}}+(-6.0 \mathrm{~N}) \hat{\mathrm{j}},\) then what is \(\vec{F}_{\text {net }}\) (a) in unit-vector notation and as (b) a magnitude and (c) an angle relative to the positive \(x\) direction? What are the (d) magnitude and (e) angle of \(\vec{a}\) ?

A \(2.00 \mathrm{~kg}\) object is subjected to three forces that give it an acceleration \(\vec{a}=-\left(8.00 \mathrm{~m} / \mathrm{s}^{2}\right) \hat{\mathrm{i}}+\left(6.00 \mathrm{~m} / \mathrm{s}^{2}\right) \hat{\mathrm{j}} . \quad\) If two of the three forces are \(\vec{F}_{1}=(30.0 \mathrm{~N}) \hat{\mathrm{i}}+(16.0 \mathrm{~N}) \hat{\mathrm{j}}\) and \(\vec{F}_{2}=\) \(-(12.0 \mathrm{~N}) \hat{\mathrm{i}}+(8.00 \mathrm{~N}) \hat{\mathrm{j}},\) find the third force.

Two horizontal forces act on a \(2.0 \mathrm{~kg}\) chopping block that can slide over a frictionless kitchen counter, which lies in an \(x y\) plane. One force is \(\vec{F}_{1}=(3.0 \mathrm{~N}) \hat{\mathrm{i}}+(4.0 \mathrm{~N}) \hat{\mathrm{j}} .\) Find the acceleration of the chopping block in unit-vector notation when the other force is (a) \(\vec{F}_{2}=(-3.0 \mathrm{~N}) \hat{\mathrm{i}}+(-4.0 \mathrm{~N}) \hat{\mathrm{j}}\) (b) \(\vec{F}_{2}=(-3.0 \mathrm{~N}) \hat{\mathrm{i}}+(4.0 \mathrm{~N}) \hat{\mathrm{j}}\) and \((\mathrm{c}) \vec{F}_{2}=(3.0 \mathrm{~N}) \hat{\mathrm{i}}+(-4.0 \mathrm{~N}) \hat{\mathrm{j}}\)

An elevator cab that weighs \(27.8 \mathrm{kN}\) moves upward. What is the tension in the cable if the cab's speed is (a) increasing at a rate of \(1.22 \mathrm{~m} / \mathrm{s}^{2}\) and \((\mathrm{b})\) decreasing at a rate of \(1.22 \mathrm{~m} / \mathrm{s}^{2} ?\)
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