/*! 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 66 A 5.00-kg package slides 2.80 m ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 66

A 5.00-kg package slides 2.80 m down a long ramp that is inclined at 24.0\(^\circ\) below the horizontal. The coefficient of kinetic friction between the package and the ramp is \(\mu_k\) \(=\) 0.310. Calculate (a) the work done on the package by friction; (b) the work done on the package by gravity; (c) the work done on the package by the normal force; (d) the total work done on the package. (e) If the package has a speed of 2.20 m/s at the top of the ramp, what is its speed after it has slid 2.80 m down the ramp?

Short Answer

Expert verified
(a) -38.92 J, (b) 55.92 J, (c) 0 J, (d) 17.0 J, (e) 3.41 m/s.

Step by step solution

01

Calculate the Force of Friction

First, determine the force of friction. The force of friction can be calculated using the formula \( f_k = \mu_k \cdot N \), where \( N \) is the normal force. On an inclined plane, \( N = mg\cos\theta \), where \( m = 5.00 \text{ kg} \), \( g = 9.81 \text{ m/s}^2 \), and \( \theta = 24.0^\circ \). Calculate: \[ N = 5.00 \cdot 9.81 \cdot \cos(24.0^\circ) \approx 44.85 \text{ N} \]\[ f_k = 0.310 \times 44.85 \approx 13.90 \text{ N} \]
02

Calculate Work Done by Friction

The work done by friction is given by \( W_f = f_k \cdot d \cdot \cos(180^\circ) \), where \( d = 2.80 \text{ m} \) is the distance. \[ W_f = 13.90 \times 2.80 \times (-1) \]\[ W_f = -38.92 \text{ J} \]
03

Calculate Work Done by Gravity

The work done by gravity is given by \( W_g = mgd\sin\theta \). Use the values \( m = 5.00 \text{ kg}, g = 9.81 \text{ m/s}^2, \theta = 24.0^\circ, d = 2.80 \text{ m} \).\[ W_g = 5.00 \times 9.81 \times 2.80 \times \sin(24.0^\circ) \approx 55.92 \text{ J} \]
04

Calculate Work Done by Normal Force

The work done by the normal force is zero because the normal force acts perpendicular to the direction of motion. Hence,\[ W_n = 0 \text{ J} \]
05

Calculate Total Work Done

The total work done on the package is the sum of the work done by friction, gravity, and the normal force.\[ W_{total} = W_f + W_g + W_n \]\[ W_{total} = -38.92 + 55.92 + 0 = 17.0 \text{ J} \]
06

Calculate Final Speed

Use the work-energy principle to find the final speed of the package. The net work done on the package is equal to the change in kinetic energy.\[ W_{total} = \frac{1}{2} mv_f^2 - \frac{1}{2} mv_i^2 \]Rearrange to find \( v_f \):\[ 17.0 = \frac{1}{2} \cdot 5.00 \cdot v_f^2 - \frac{1}{2} \cdot 5.00 \cdot 2.20^2 \]Solve for \( v_f \):\[ 17.0 = \frac{1}{2} \cdot 5.00 \cdot (v_f^2 - 4.84) \]\[ 17.0 = 2.50(v_f^2 - 4.84) \]\[ 6.8 = v_f^2 - 4.84 \]\[ v_f^2 = 11.64 \]\[ v_f = \sqrt{11.64} \approx 3.41 \text{ m/s} \]

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 91影视!

Key Concepts

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

Kinetic Friction
Kinetic friction is a force that opposes the motion of objects sliding against each other. It plays a crucial role in slowing things down. When an object moves across a surface, kinetic friction acts in the opposite direction of the object's movement.
In our exercise, the coefficient of kinetic friction (\( \mu_k = 0.310 \)) tells us how "sticky" the two surfaces are. The normal force (\( N \)) presses the object against the surface. It depends on the object's weight and the angle of the slope.To calculate the force of kinetic friction, we use the equation:
  • \( f_k = \mu_k \times N \)
This frictional force results in energy loss, affecting how objects move over time.
In scenarios like our inclined plane, understanding kinetic friction is essential to predict the motion and final speed of the object.
Inclined Plane
An inclined plane is a flat surface tilted at an angle, and it's a simple machine used to make lifting objects easier. In physics problems, an inclined plane helps demonstrate how gravity, friction, and other forces interact.
When an object is on an inclined plane, the gravitational force can be split into two components:
  • One parallel to the plane, causing the object to slide down.
  • Another perpendicular to the plane, acting as the normal force.
For our 24.0-degree incline, the parallel component is calculated using \( mg\sin(\theta) \) and the perpendicular using \( mg\cos(\theta) \).These components allow us to determine how the object moves down the plane and how friction affects that motion. Analyzing such setups helps us study complex real-world interactions in a simpler manner.
Net Work
In physics, work is done when a force moves an object over a distance. Net work is the total work done by all forces on an object. It's vital in understanding how energy transfers between forms and how it affects an object's motion.
When calculating net work, consider contributions from all forces:
  • Work done by gravity.
  • Work done by friction (often negative since it opposes motion).
  • Work by normal force (usually zero as it's perpendicular to movement).
In our scenario, the net work done is:\( W_{total} = W_f + W_g + W_n \)This net work translates into a change in kinetic energy, as described by the work-energy principle. This principle is key to predicting how the object's speed changes after moving a certain distance. By adding or subtracting energy through work, we can determine the final velocity of the package sliding down the ramp.

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

An ingenious bricklayer builds a device for shooting bricks up to the top of the wall where he is working. He places a brick on a vertical compressed spring with force constant \(k = 450\) N/m and negligible mass. When the spring is released, the brick is propelled upward. If the brick has mass 1.80 kg and is to reach a maximum height of 3.6 m above its initial position on the compressed spring, what distance must the bricklayer compress the spring initially? (The brick loses contact with the spring when the spring returns to its uncompressed length. Why?)

The human heart is a powerful and extremely reliable pump. Each day it takes in and discharges about 7500 L of blood. Assume that the work done by the heart is equal to the work required to lift this amount of blood a height equal to that of the average American woman (1.63 m). The density (mass per unit volume) of blood is \(1.05 \times 10^3 \, \mathrm{kg/m}^3\). (a) How much work does the heart do in a day? (b) What is the heart's power output in watts?

All birds, independent of their size, must maintain a power output of 10\(-\)25 watts per kilogram of body mass in order to fly by flapping their wings. (a) The Andean giant hummingbird (\(Patagona gigas\)) has mass 70 g and flaps its wings 10 times per second while hovering. Estimate the amount of work done by such a hummingbird in each wingbeat. (b) A 70-kg athlete can maintain a power output of 1.4 kW for no more than a few seconds; the \(steady\) power output of a typical athlete is only 500 W or so. Is it possible for a human-powered aircraft to fly for extended periods by flapping its wings? Explain.

When its 75-kW (100-hp) engine is generating full power, a small single-engine airplane with mass 700 kg gains altitude at a rate of 2.5 m/s (150 m/min, or 500 ft/min). What fraction of the engine power is being used to make the airplane climb? (The remainder is used to overcome the effects of air resistance and of inefficiencies in the propeller and engine.)

It is 5.0 km from your home to the physics lab. As part of your physical fitness program, you could run that distance at 10 km/h (which uses up energy at the rate of 700 W), or you could walk it leisurely at 3.0 km/h (which uses energy at 290 W). Which choice would burn up more energy, and how much energy (in joules) would it burn? Why does the more intense exercise burn up less energy than the less intense exercise?
See all solutions

Recommended explanations on Physics Textbooks

Linear Momentum

Read Explanation

Dynamics

Read Explanation

Work Energy and Power

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation

Fluids

Read Explanation

Scientific Method Physics

Read Explanation
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.