/*! 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 41 You are working for a shipping c... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

You are working for a shipping company. Your job is to stand at the bottom of an 8.0 -m-long ramp that is inclined at \(37^{\circ}\) above the horizontal. You grab packages off a conveyor belt and propel them up the ramp. The coefficient of kinetic friction between the packages and the ramp is \(\mu_{\mathrm{k}}=0.30 .\) (a) What speed do you need to give a package at the bottom of the ramp so that it has zero speed at the top of the ramp? (b) Your coworker is supposed to grab the packages as they arrive at the top of the ramp, but she misses one and it slides back down. What is its speed when it returns to you?

Short Answer

Expert verified
(a) The initial speed is 6.5 m/s. (b) The return speed is 4.2 m/s.

Step by step solution

01

Identify the known variables

We have a ramp of length 8.0 m inclined at an angle of \( 37^{\circ} \). The coefficient of kinetic friction \( \mu_k \) is 0.30. We need to find the initial speed needed and the speed when the package returns.
02

Analyze energy changes for part (a)

We will use the principle of conservation of energy where the initial kinetic energy and work done against friction equal the potential energy at the top. Formula: \[ K_i - W_{friction} = U_f \] where \( K_i = \frac{1}{2} m v^2 \), \( W_{friction} = f_k d \) and \( U_f = mgh \).
03

Find expressions for energy components

Calculate the work done by friction: \[ f_k = \mu_k \cdot m \cdot g \cdot \cos(37^{\circ}) \] Work done by friction: \[ W_{friction} = f_k \cdot d \] Potential energy at the top: \[ U_f = m \cdot g \cdot h \] where \( h = d \cdot \sin(37^{\circ}) \).
04

Solve for initial velocity for (a)

Setting up the energy equation: \[ \frac{1}{2}mv^2 = mgh + \mu_k \cdot m \cdot g \cdot \cos(37^{\circ}) \cdot d \] Cancel \( m \) and solve for \( v \): \[ v^2 = 2gh + 2\mu_k \cdot g \cdot \cos(37^{\circ}) \cdot d \] Substitute known values to find \( v \).
05

Calculate initial speed for (a)

Substitute \( g = 9.8 \, \text{m/s}^2 \), \( h = 8 \cdot \sin(37^{\circ}) \), and \( d = 8 \) m:\[ v^2 = 2 \times 9.8 \times 8 \times \sin(37^{\circ}) + 2 \times 0.30 \times 9.8 \times 8 \times \cos(37^{\circ}) \] Calculate to find \( v = 6.5 \, \text{m/s} \).
06

Analyze energy changes for part (b)

The package starts from rest at the top of the ramp and slides back down. Use energy conservation again: Potential energy at top equals kinetic energy at the bottom plus work done by friction: \[ mgh = \frac{1}{2} mv^2 + \mu_k m g d \cos(37^{\circ}) \]
07

Solve for speed at bottom for (b)

Cancel \( m \) and rearrange to find \( v \):\[ \frac{1}{2}v^2 = gh - \mu_k \cdot g \cdot d \cdot \cos(37^{\circ}) \] Substitute known values to find \( v \).
08

Calculate final speed for (b)

Substitute \( g = 9.8 \, m/s^2 \) and given values:\[ \frac{1}{2}v^2 = 9.8 \times 8 \times \sin(37^{\circ}) - 0.30 \times 9.8 \times 8 \times \cos(37^{\circ}) \] Calculate to find \( v = 4.2 \, \text{m/s} \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Conservation of Energy
The principle of conservation of energy is essential for solving problems where energy is transformed from one type to another. When dealing with physics problems like propelling a package up an inclined plane, this principle states that the total mechanical energy of the package is conserved if we ignore any external energy input or loss, except gravity and friction.
Energy can exist in several forms such as kinetic energy (energy of motion) and potential energy (energy stored due to position). For our inclined plane problem, this means that the kinetic energy you impart to the package at the bottom of the ramp converts into potential energy as it reaches the top. There is also some energy lost due to friction, which needs to be considered.
  • Initial Kinetic Energy: The energy given to the package as it starts its upward journey. Expressed as \( \frac{1}{2}mv^2 \).
  • Work done by Friction: Energy lost as the package slides up, calculated using the frictional force \( f_k \) and the distance \( d \).
  • Potential Energy at the Top: The energy the package has when it reaches the height, expressed as \( mgh \).
In our problem, you'll use the conservation of energy principle to find the speed required to just reach the top with zero speed and also the speed when it returns to the bottom.
Inclined Plane
An inclined plane is a flat surface tilted at an angle to the horizontal. It allows objects to be moved with less effort compared to lifting them vertically. When dealing with inclined planes in physics problems, it's important to analyze how gravity and other forces interact with the object on the slope.
The ramp in our problem is inclined at a specific angle, which means the gravitational force acting on the package needs to be broken into components:
  • Parallel component: The component of gravity that acts down the slope, trying to accelerate the package down. It is given by \( mg\sin(\theta) \).
  • Perpendicular component: The component acting into the slope, balancing the normal force and given by \( mg\cos(\theta). \)
These components are crucial for calculating both the work done against friction and the changes in energy. By understanding how these forces work, you can solve for the initial speed required for the package to reach the top.
Friction
Friction is a force that opposes the motion of an object. It acts in the opposite direction to the object's movement. In our inclined plane problem, kinetic friction plays a critical role. It needs to be overcome for the package to ascend the ramp.
Kinetic friction depends on two main factors:
  • Coefficient of Kinetic Friction \( \mu_k \): A dimensionless number that represents the friction between the surfaces. Given as 0.30 in the exercise.
  • Normal Force \( N \): The perpendicular force between the package and the slope, calculated using \( mg\cos(\theta) \).
The force of friction \( f_k \) is computed as \( \mu_k \cdot N \). This force is then used to calculate the work done by friction \( W_{friction} = f_k \cdot d \), which must be considered in energy calculations to find the initial and final speeds in both parts of the problem.
Physics Problem Solving
Solving physics problems involves a systematic approach to dissecting and analyzing the situation. Here’s a simple approach that applies to the incline plane problem:
  • Identify Known Variables: Determine what's given in the problem, such as ramp length, angle, coefficient of friction, and gravity.
  • Establish Equations: Use relevant physics equations, like the conservation of energy, to set up relationships between known and unknown quantities.
  • Handle Units Carefully: Ensure all units are consistent, typically in meters, kilograms, and seconds.
  • Substitute and Solve: Insert the known values and solve the equations for the requested unknown variables, step by step, scrupulously checking your work.
By methodically breaking down each facet of the problem, such as identifying energy transfer or friction effects, you can reliably reach the correct solution. Remember, practice in breaking down complex problems into simpler parts can greatly enhance your problem-solving skills.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

One straightforward way to measure the coefficients of friction between a box and a wooden surface is illustrated in Figure \(5.59 .\) The sheet of wood can be raised by pivoting it about one edge. It is first raised to an angle \(\theta_{1}\) (which is measured) for which the box just begins to slide downward. The sheet is then immediately lowered to an angle \(\theta_{2}\) (which is also measured) for which the box slides with constant speed down the sheet. Apply Newton's second law to the box in both cases to find the coefficients of kinetic and static friction between it and the wooden sheet in terms of the measured angles \(\theta_{1}\) and \(\theta_{2}\).

An \(80 \mathrm{~N}\) box initially at rest is pulled by a horizontal rope on a horizontal table. The coefficients of kinetic and static friction between the box and the table are \(\frac{1}{4}\) and \(\frac{1}{2},\) respectively. What is the friction (b) \(25 \mathrm{~N}\), force on this box if the tension in the rope is (a) \(0 \mathrm{~N},]\) (c) \(39 \mathrm{~N},(\mathrm{~d}) 41 \mathrm{~N},(\mathrm{e}) 150 \mathrm{~N} ?\)

A person pushes on a stationary \(125 \mathrm{~N}\) box with \(75 \mathrm{~N}\) at \(30^{\circ}\) below the horizontal, as shown in Figure 5.61 . The coefficient of static friction between the box and the horizontal floor is 0.80 . (a) Make a free-body diagram of the box. (b) What is the normal force on the box? (c) What is the friction force on the box? (d) What is the largest the friction force could be? (e) The person now replaces his push with a \(75 \mathrm{~N}\) pull at \(30^{\circ}\) above the horizontal. Find the normal force on the box in this case.

In emergencies involving major blood loss, the doctor will order the patient placed in the Trendelberg position, which is to raise the foot of the bed to get maximum blood flow to the brain. If the coefficient of static friction between the typical patient and the bedsheets is \(1.2,\) what is the maximum angle at which the bed can be tilted with respect to the floor before the patient begins to slide?

An astronaut on the distant planet Xenon uses an adjustable inclined plane to measure the acceleration of gravity. The plane is frictionless, and its angle of inclination can be varied. Here is a table of the data: $$\begin{array}{lc}\hline \theta & a\left(\mathrm{~m} / \mathrm{s}^{2}\right) \\\\\hline 5.0^{\circ} & 1.20 \\ 10^{\circ} & 2.49 \\\15^{\circ} & 3.59 \\\20^{\circ} & 4.90 \\\25^{\circ} & 5.95 \\\\\hline\end{array}$$ Make a plot of the measured acceleration as a function of the sine of the angle of incline. Using a linear "best fit" to the data, determine the value of \(g\) on the planet Xenon.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.