/*! 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 84 You pull a short refrigerator wi... [FREE SOLUTION] | 91影视

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Chapter 5: Problem 84

You pull a short refrigerator with a constant force \(\vec{F}\) across a greased (frictionless) floor, either with \(F\) horizontal (case 1 ) or with \(\vec{F}\) tilted upward at an angle \(\theta\) (case 2 ). (a) What is the ratio of the refrigerator's speed in case 2 to its speed in case 1 if you pull for a certain time \(t ?\) (b) What is this ratio if you pull for a certain distance \(d ?\)

Short Answer

Expert verified
(a) Ratio is \( \cos \theta \). (b) Ratio is \( \sqrt{\cos \theta} \).

Step by step solution

01

Understanding the Problem

We need to find the ratio of the refrigerator's speed in two different scenarios. In case 1, the force is applied horizontally, and in case 2, the force is applied at an angle \( \theta \) upwards. The scenarios are evaluated for a given time \( t \) in part (a) and a given distance \( d \) in part (b).
02

Calculating Final Speed for Certain Time (Part a)

In both cases, because there is no friction, the net force is the component of \( \vec{F} \) parallel to the floor. For case 1, the entire force is horizontal: \( F_1 = F \). For case 2, we take the horizontal component of \( \vec{F} \): \( F_2 = F \cos\theta \). The acceleration in each case is \( a = F/m \). Thus, final speeds are \( v_1 = (F/m) t \) and \( v_2 = (F\cos\theta/m) t \).
03

Calculating Speed Ratio for Certain Time

The ratio of speeds when pulled for a certain time \( t \) is calculated as: \( \frac{v_2}{v_1} = \frac{(F \cos \theta/m) t}{(F/m) t} = \cos \theta \).
04

Calculating Final Speed for Certain Distance (Part b)

When focusing on a certain distance, the work-energy principle applies. The work done by force equals the change in kinetic energy. For case 1, \( W_1 = Fd \) and for case 2, the work is \( W_2 = Fd \cos \theta \). The relation of velocity with work is \( v^2 = 2(W/m) \). Thus, \( v_1^2 = 2(Fd/m) \) and \( v_2^2 = 2(Fd \cos \theta/m) \).
05

Calculating Speed Ratio for Certain Distance

The speed ratio for a certain distance \( d \) is: \( \frac{v_2}{v_1} = \sqrt{\frac{2(Fd \cos \theta/m)}{2(Fd/m)}} = \sqrt{\cos \theta} \).
06

Conclusion and Final Ratios

After solving both parts, the ratio of the refrigerator's speed when moving for a certain time is \( \cos \theta \), and for a certain distance, it is \( \sqrt{\cos \theta} \).

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Key Concepts

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

Frictionless Motion
In a frictionless environment, objects move without resistance from surfaces around them. This means there are no forces opposing the motion. As you pull the refrigerator across the greased floor, it slides without any hindrance from friction.
This simplifies physics problems because we do not need to account for frictional forces, which would otherwise slow down the object.
  • No energy is lost to heat due to friction.
  • The motion is determined solely by the applied force.
This concept is crucial for understanding how the refrigerator moves when you apply a force, as it allows us to focus on other variables like force direction and time.
Force Components
When we deal with forces at angles, it's helpful to break them down into components. These components are parts of the force that act in specific directions 鈥� usually horizontal and vertical.
This is particularly applicable in case 2 of our problem, where the force isn鈥檛 applied directly horizontally, but at an angle \(\theta\).
  • Horizontal Component: Given by \(F \cos \theta\).
  • Vertical Component: Given by \(F \sin \theta\).
It's the horizontal component, \(F \cos \theta\), that contributes to the forward motion of the refrigerator on the horizontal floor. By understanding force components, you can more intuitively predict how an angled force will affect motion.
Kinematics
Kinematics deals with the motion of objects without considering the forces that cause this motion. In our exercise, we calculate speeds and how they change over time or distance.
For the refrigerator moving across a frictionless floor:
  • Speed for certain time \(t\): We use the formula \(v = \frac{F}{m} t\) for consistent acceleration, driven by the continuous force \(F\).
  • Speed for certain distance \(d\): We use the relation \(v^2 = 2 \left(\frac{W}{m}\right)\) which connects work done to change in kinetic energy.
Understanding kinematics allows us to predict and describe how the refrigerator's speed will vary, given different scenarios of force application over time or distance.
Work-Energy Principle
The work-energy principle helps us understand how the work done by forces affects an object's energy. When you apply force to move an object, you do work on it, and this work translates into the object's kinetic energy.
The formula for work is \(W = F \times d\) for force applied over a distance \(d\). In our refrigerator scenario, this principle connects the work done to the change in the refrigerator's speed.
  • Case 1 (Horizontal Force): Work is simply \(Fd\).
  • Case 2 (Angled Force): Work is reduced by \(\cos \theta\), giving \(Fd \cos \theta\), because only part of the force contributes to movement.
Using the work-energy principle, we relate velocity and work directly. This gives us an understanding of how the differing amounts of work done in each case affect the velocities.

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Most popular questions from this chapter

An \(85 \mathrm{~kg}\) man lowers himself to the ground from a height of \(10.0 \mathrm{~m}\) by holding onto a rope that runs over a frictionless pulley to a \(65 \mathrm{~kg}\) sandbag. With what speed does the man hit the ground if he started from rest?

An object is hung from a spring balance attached to the ceiling of an elevator cab. The balance reads \(65 \mathrm{~N}\) when the cab is standing still. What is the reading when the cab is moving upward (a) with a constant speed of \(7.6 \mathrm{~m} / \mathrm{s}\) and \((\mathrm{b})\) with a speed of \(7.6 \mathrm{~m} / \mathrm{s}\) while decelerating at a rate of \(2.4 \mathrm{~m} / \mathrm{s}^{2} ?\)

The high-speed winds around a tornado can drive projectiles into trees, building walls, and even metal traffic signs. In a laboratory simulation, a standard wood toothpick was shot by pneumatic gun into an oak branch. The toothpick's mass was \(0.13 \mathrm{~g}\), its speed before entering the branch was \(220 \mathrm{~m} / \mathrm{s},\) and its penetration depth was \(15 \mathrm{~mm} .\) If its speed was decreased at a uniform rate, what was the magnitude of the force of the branch on the toothpick?

A customer sits in an amusement park ride in which the compartment is to be pulled downward in the negative direction of a \(y\) axis with an acceleration magnitude of \(1.24 g,\) with \(g=9.80 \mathrm{~m} / \mathrm{s}^{2}\). A \(0.567 \mathrm{~g}\) coin rests on the customer's knee. Once the motion begins and in unit-vector notation, what is the coin's acceleration relative to (a) the ground and (b) the customer? (c) How long does the coin take to reach the compartment ceiling, \(2.20 \mathrm{~m}\) above the knee? In unit-vector notation, what are (d) the actual force on the coin and (e) the apparent force according to the customer's measure of the coin's acceleration?

An elevator cab and its load have a combined mass of \(1600 \mathrm{~kg}\). Find the tension in the supporting cable when the cab, originally moving downward at \(12 \mathrm{~m} / \mathrm{s}\), is brought to rest with constant acceleration in a distance of \(42 \mathrm{~m}\).
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