/*! 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 65 CP A 20.0 -kg crate sits at rest... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 65

CP A 20.0 -kg crate sits at rest at the bottom of a 15.0 -m-long ramp that is inclined at \(34.0^{\circ}\) above the horizontal. A constant horizontal force of 290 \(\mathrm{N}\) is applied to the crate to push it up the ramp. While the crate is moving, the ramp exerts a constant frictional force on it that has magnitude 65.0 \(\mathrm{N}\) . (a) What is the total work done on the crate during its motion from the bottom to the top of the ramp? (b) How much time does it take the crate to travel to the top of the ramp?

Short Answer

Expert verified
(a) The total work done is determined by calculating net force and multiplying by the ramp length. (b) Use kinematics to find the time given the acceleration.

Step by step solution

01

Analyze the forces on the crate

Identify the forces acting on the crate: the applied horizontal force (290 N), the gravitational force acting downward, the frictional force (65.0 N), and the normal force perpendicular to the ramp's surface. The gravitational force can be calculated as \( F_g = mg \), where \( m = 20.0 \, \text{kg} \) and \( g = 9.81 \, \text{m/s}^2 \).
02

Determine the component of the applied force along the ramp

The applied force of 290 N is horizontal, but we need to find the component along the ramp: \( F_{app, \text{parallel}} = F_{app} \cdot \cos(\theta) \), where \( \theta = 34.0^{\circ} \). Compute this component to use in further calculations.
03

Calculate the gravitational force component parallel to the ramp

The component of gravitational force acting along the ramp is \( F_{g, \text{parallel}} = mg \cdot \sin(\theta) \). Use \( m = 20.0 \, \text{kg} \) and \( \theta = 34.0^{\circ} \) to find \( F_{g, \text{parallel}} \).
04

Find the net force along the ramp

The net force along the ramp, \( F_{net} \), is given by the sum of forces along the ramp: \( F_{net} = F_{app, \text{parallel}} - F_{g, \text{parallel}} - F_{friction} \), where \( F_{friction} = 65.0 \, \text{N} \). Calculate \( F_{net} \).
05

Calculate the work done by the net force

The work done by the net force is given by \( W = F_{net} \cdot d \), where \( d = 15.0 \, \text{m} \) is the length of the ramp. Substitute the value of \( F_{net} \) to find the total work done on the crate.
06

Calculate the acceleration of the crate

Using Newton's second law, \( F_{net} = ma \), solve for the acceleration \( a \) of the crate, where \( m = 20.0 \, \text{kg} \).
07

Determine the time taken to reach the top of the ramp

Use the kinematic equation \( d = \frac{1}{2} a t^2 \) to solve for the time \( t \), given \( d = 15.0 \, \text{m} \) and the calculated acceleration \( a \).

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.

Force Analysis
Understanding the forces acting on an object is crucial in physics. In our scenario involving the crate, several forces are at play:
  • Applied Force: A horizontal force of 290 N pushes the crate up the ramp.
  • Gravitational Force: This force acts downward and is calculated as the product of mass (20.0 kg) and acceleration due to gravity (9.81 m/s虏), resulting in a gravitational force downward.
  • Frictional Force: Acts in the opposite direction to the crate's movement, with a magnitude of 65.0 N.
  • Normal Force: Acts perpendicular to the ramp's surface, balancing the component of the gravitational force perpendicular to the ramp.
To solve related problems, consider both the direction and magnitude of these forces. Force analysis involves breaking down forces into components, especially in inclined motion, to understand their contribution to net force.
Frictional Force
Frictional forces are a core element when considering objects moving across surfaces. It is the resistance force that opposes the relative motion of an object. In the case of the crate moving up the inclined ramp, friction acts to slow it down. The frictional force here is a constant 65.0 N, opposing the motion. Frictional forces depend on:
  • The nature of surfaces in contact.
  • The normal force pressing the surfaces together. This is perpendicular to the direction of motion and impacts the frictional force significantly.
When calculating net force, always include the frictional force as it reduces the effectiveness of the applied force, thus impacting acceleration and travel time.
Kinematics
Kinematics is the study of motion without regarding the forces causing it. For the crate, we are interested in how it moves up the ramp:Using kinematic equations, we can determine how long it takes to reach the top:
  • The primary equation here is: \[ d = \frac{1}{2} a t^2 \]where \(d\) is the distance (15.0 m), \(a\) is the acceleration obtained from force analysis, and \(t\) is the time to find.
  • Solve this equation for \(t\) to find the duration of travel.
Understanding these principles helps predict motion accurately and interpret the effects of different forces acting on the crate.
Newton's Laws
Newton's Laws of Motion are fundamental to understanding force and motion.- First Law (Inertia): States that an object will remain at rest or in uniform motion unless acted on by an external force. Since the crate is initially at rest, a force is necessary to push it up the ramp.- Second Law (F=ma): Relates force, mass, and acceleration. We use this law to calculate the crate鈥檚 acceleration: \[ F_{net} = ma \]Net force (\(F_{net}\)) is the sum of all forces acting along the ramp, solving for acceleration \(a\).- Third Law (Action-Reaction): Every action has an equal and opposite reaction. The crate exerts a force on the ramp, and the ramp exerts an equal force in the opposite direction.Understanding these laws aids in developing a comprehensive view of motion and forces, enhancing problem-solving skills in physics.

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

CPA small block with a mass of 0.0900 \(\mathrm{kg}\) is attached to a cord passing through a hole in a frictionless, horizontal surface (Fig. P6.75). The block is originally revolving at a distance of 0.40 \(\mathrm{m}\) from the hole with a speed of 0.70 \(\mathrm{m} / \mathrm{s} .\) The cord is then pulled from below, shortening the radius of the circle in which the block revolves to 0.10 \(\mathrm{m} .\) At this new distance, the speed of the block is observed to be 2.80 \(\mathrm{m} / \mathrm{s}\) . (a) What is the tension in the cord in the original situation when the block has speed \(v=0.70 \mathrm{m} / \mathrm{s} ?\) (b) What is the tension in the cord in the final situation when the block has speed \(v=2.80 \mathrm{m} / \mathrm{s} ?\) (c) How much work was done by the person who pulled on the cord?

Use the work-energy theorem to solve each of these problems. You can use Newton's laws to check your answers. Neglect air resistance in all cases. (a) A branch falls from the top of a 95.0 -m-tall redwood tree, starting from rest. How fast is it moving when it reaches the ground? (b) A volcano ejects a boulder directly upward 525 \(\mathrm{m}\) into the air. How fast was the boulder moving just as it left the volcano? (c) A skier moving at 5.00 \(\mathrm{m} / \mathrm{s}\) encounters a long, rough horizontal patch of snow having coefficient of kinetic friction 0.220 with her skis. How far does she travel on this patch before stopping? (d) Suppose the rough patch in part (c) was only 2.90 m long? How fast would the skier be moving when she reached the end of the patch? (e) At the base of a frictionless icy hill that rises at \(25.0^{\circ}\) above the horizontal, a toboggan has a speed of 12.0 \(\mathrm{m} / \mathrm{s}\) toward the hill. How high vertically above the base will it go before stopping?

At a waterpark, sleds with riders are sent along a slippery, horizontal surface by the release of a large compressed spring. The spring with force constant \(k=40.0 \mathrm{N} / \mathrm{cm}\) and negligible mass rests on the frictionless horizontal surface. One end is in contact with a stationary wall. A sled and rider with total mass 70.0 \(\mathrm{kg}\) are pushed against the other end, compressing the spring 0.375 \(\mathrm{m}\) . The sled is then released with zero initial velocity. What is the sled's speed when the spring (a) returns to its uncompressed length and (b) is still compressed 0.200 \(\mathrm{m} ?\)

You are a member of an Alpine Rescue Team. You must project a box of supplies up an incline of constant slope angle \(\alpha\) so that it reaches a stranded skier who is a vertical distance \(h\) above the bottom of the incline. The incline is slippery, but there is some friction present, with kinetic friction coefficient \(\mu_{\mathrm{k}}.\) Use the work- energy theorem to calculate the minimum speed you must give the box at the bottom of the incline so that it will reach the skier. Express your answer in terms of \(g, h, \mu_{\mathrm{k}},\) and \(\alpha\).

BIO Power of the Human Heart. The human heart is a powerful and extremely reliable pump. Each day it takes in and discharges about 7500 \(\mathrm{L}\) 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 \((1.63 \mathrm{m}) .\) The density (mass per unit volume) of blood is \(1.05 \times 10^{3} \mathrm{kg} / \mathrm{m}^{3}\) . (a) How much work does the heart do in a day? (b) What is the heart's power output in watts?
See all solutions

Recommended explanations on Physics Textbooks

Dynamics

Read Explanation

Rotational Dynamics

Read Explanation

Nuclear Physics

Read Explanation

Electrostatics

Read Explanation

Atoms and Radioactivity

Read Explanation

Radiation

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.