/*! 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 94 A 10.0 -kg block is released fro... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 94

A 10.0 -kg block is released from rest on a frictionless track inclined at an angle of \(55^{\circ} .\) (a) What is the net force on the block after it is released? (b) What is the acceleration of the block? (c) If the block is released from rest, how long will it take for the block to attain a speed of \(10.0 \mathrm{m} / \mathrm{s} ?\) (d) Draw a motion diagram for the block. (e) Draw a graph of \(v_{x}(t)\) for values of velocity between 0 and $10 \mathrm{m} / \mathrm{s}\(. Let the positive \)x$ -axis point down the track.

Short Answer

Expert verified
Answer: The net force acting on the block is 80.0 N, and its acceleration is 8.0 m/s虏. It takes 1.25 seconds for the block to reach a velocity of 10.0 m/s.

Step by step solution

01

Finding the gravitational force acting on the block

First, find the gravitational force acting on the block. This is given by: $$ F_g = m \times g $$ where \(m=10.0\,\text{kg}\) is the mass of the block, and \(g=9.81\,\text{m/s}^2\) is the acceleration due to gravity. Substituting the values, we get: $$ F_g = 10.0\,\text{kg} \times 9.81\,\text{m/s}^2 = 98.1\,\text{N} $$
02

Decomposing the gravitational force into components

We can decompose the gravitational force acting on the block into two components: one parallel and one perpendicular to the inclined surface. Given the angle of inclination \(\theta=55^{\circ}\), $$ F_{gx} = F_g \sin\theta \quad \text{and}\quad F_{gy} = F_g \cos\theta $$ Plugging in the calculated \(F_g=98.1\,\text{N}\) and given angle \(\theta=55^{\circ}\), we get: $$ F_{gx} = 98.1\,\text{N} \sin 55^{\circ} \approx 80.0\,\text{N} $$ $$ F_{gy} = 98.1\,\text{N} \cos 55^{\circ} \approx 56.2\,\text{N} $$
03

Newton's second law of motion

Since there is no external horizontal force acting on the block, using Newton's second law of motion in the x-direction (parallel to the inclined plane) yields: $$ F_{net} = m a_x $$ Using the calculated value for \(F_{gx}=80.0\, \text{N}\), $$ a_x = \frac{F_{gx}}{m} = \frac{80.0\,\text{N}}{10.0\,\text{kg}} = 8.0\,\text{m/s}^2 $$ The net force \(F_{net}\) is equal to \(F_{gx}\), so the net force is \(80.0\,\text{N}\), and the acceleration of the block is \(8.0\,\text{m/s}^2\).
04

Time to reach a velocity of 10.0 m/s

Using the kinematic equation for velocity: $$ v_x = v_{0x} + a_x t $$ where \(v_x\) is the final velocity, \(v_{0x}\) is the initial velocity, \(a_x\) is the acceleration, and \(t\) is the time taken to reach the final velocity. Since the block is released from rest, \(v_{0x} = 0\). We want to find \(t\) for \(v_x=10.0\,\text{m/s}\). $$ 10.0\,\text{m/s} = 0 + 8.0\,\text{m/s}^2 \times t $$ Solving for \(t\), we get: $$ t = \frac{10.0\,\text{m/s}}{8.0\,\text{m/s}^2} = 1.25\,\text{s} $$
05

Motion diagram and v_x(t) graph

(d) In the motion diagram, we can represent the block with a series of arrows. The length of each arrow is proportional to the distance moved and the horizontal distance between arrows represents equal time intervals. The arrows start from rest and gradually become longer, as the block gains speed and moves down the inclined plane. (e) To plot the graph of \(v_x(t)\) for velocities between \(0\,\text{m/s}\) and \(10\,\text{m/s}\), use the equation: $$ v_x(t)=8.0t $$ The graph is a straight line with a slope of \(8.0\,\text{m/s}^2\) and intersects the time axis at \(t=0\) (initially at rest). The velocity increases linearly with respect to time and reaches \(10.0\,\text{m/s}\) at \(t=1.25\,\text{s}\).

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!

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

A truck is towing a 1000 -kg car at a constant speed up a hill that makes an angle of \(\alpha=5.0^{\circ}\) with respect to the horizontal. A rope is attached from the truck to the car at an angle of \(\beta=10.0^{\circ}\) with respect to horizontal. Ignore any friction in this problem. (a) Draw an FBD showing all the forces on the car. Indicate the angle that each force makes with either the vertical or horizontal direction. (b) What is the tension in the rope?
The vertical component of the acceleration of a sailplane is zero when the air pushes up against its wings with a force of \(3.0 \mathrm{kN}\). (a) Assuming that the only forces on the sailplane are that due to gravity and that due to the air pushing against its wings, what is the gravitational force on the Earth due to the sailplane? (b) If the wing stalls and the upward force decreases to \(2.0 \mathrm{kN}\) what is the acceleration of the sailplane?
During a balloon ascension, wearing an oxygen mask, you measure the weight of a calibrated \(5.00-\mathrm{kg}\) mass and find that the value of the gravitational field strength at your location is 9.792 N/kg. How high above sea level, where the gravitational field strength was measured to be $9.803 \mathrm{N} / \mathrm{kg},$ are you located?
A model sailboat is slowly sailing west across a pond at $0.33 \mathrm{m} / \mathrm{s} .\( A gust of wind blowing at \)28^{\circ}$ south of west gives the sailboat a constant acceleration of magnitude $0.30 \mathrm{m} / \mathrm{s}^{2}\( during a time interval of \)2.0 \mathrm{s} .$ (a) If the net force on the sailboat during the 2.0 -s interval has magnitude $0.375 \mathrm{N},$ what is the sailboat's mass? (b) What is the new velocity of the boat after the 2.0 -s gust of wind?
Two canal workers pull a barge along the narrow waterway at a constant speed. One worker pulls with a force of \(105 \mathrm{N}\) at an angle of \(28^{\circ}\) with respect to the forward motion of the barge and the other worker, on the opposite tow path, pulls at an angle of \(38^{\circ}\) relative to the barge motion. Both ropes are parallel to the ground. (a) With what magnitude force should the second worker pull to make the sum of the two forces be in the forward direction? (b) What is the magnitude of the force on the barge from the two tow ropes?
See all solutions

Recommended explanations on Physics Textbooks

Magnetism

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation

Waves Physics

Read Explanation

Medical Physics

Read Explanation

Fundamentals of Physics

Read Explanation

Oscillations

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.