/*! 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 93 In the physics laboratory, a gli... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 93

In the physics laboratory, a glider is released from rest on a frictionless air track inclined at an angle. If the glider has gained a speed of $25.0 \mathrm{cm} / \mathrm{s}\( in traveling \)50.0 \mathrm{cm}$ from the starting point, what was the angle of inclination of the track? Draw a graph of \(v_{x}(t)\) when the positive \(x\) -axis points down the track.

Short Answer

Expert verified
The angle of inclination of the track is approximately \(0.447^{\circ}\). The graph of \(v_{x}(t)\) is a straight line starting at the origin, with a slope of \(0.125\).

Step by step solution

01

Convert cm to m and setup known variables

To make the calculations easier, convert the given values into meters: Initial speed \((u) = 0\) (since the glider is released from rest) Final speed \((v) = 25.0cm/s = 0.25m/s\) Distance \((s) = 50.0cm = 0.5m\)
02

Find the acceleration along the track

We can use the equation of motion (which applies to constant acceleration) to find the acceleration along the track: $$ v^{2} = u^{2} + 2as $$ Plug in the known values and solve for \(a\): $$ (0.25)^{2} = 0^{2} + 2a(0.5) $$ $$ a = 0.125 m/s^{2} $$
03

Determine the gravitational force component along the track

Since the glider experiences gravity and the only force acting on it is the gravitational force component along the track, we have: $$ F = ma $$ $$ mg \sin\theta = ma $$ Where \(g\) is the gravitational acceleration \((9.81 m/s^{2})\) and \(\theta\) is the angle of inclination. We can now solve for \(\theta\): $$ \sin\theta = \frac{a}{g} $$
04

Calculate the angle of inclination

Plug in the values of \(a\) and \(g\) and solve for \(\theta\): $$ \sin\theta = \frac{0.125}{9.81} $$ $$ \theta = \arcsin(\frac{0.125}{9.81}) $$ $$ \theta \approx 0.447^{\circ} $$
05

Draw a graph of \(v_{x}(t)\) along the track

Use the equation of motion along the x-axis to express the final speed as a function of time for a positive x-axis pointing down the track: $$ v_{x}(t) = u + at $$ As the glider starts from rest, we have: $$ v_{x}(t) = at $$ To plot \(v_{x}(t)\), just remember that it's a straight line with the slope equal to the acceleration component along the x-axis \((a)\). The line starts at the origin (\((0,0)\)) and has a slope of \(0.125\). In summary, the angle of inclination of the track is approximately \(0.447^{\circ}\), and the graph of \(v_{x}(t)\) is a straight line starting at the origin and with a slope of \(0.125\).

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 crate of potatoes of mass \(18.0 \mathrm{kg}\) is on a ramp with angle of incline \(30^{\circ}\) to the horizontal. The coefficients of friction are \(\mu_{\mathrm{s}}=0.75\) and \(\mu_{\mathrm{k}}=0.40 .\) Find the frictional force (magnitude and direction) on the crate if the crate is at rest.
The coefficient of static friction between block \(\mathrm{A}\) and a horizontal floor is 0.45 and the coefficient of static friction between block \(\mathrm{B}\) and the floor is \(0.30 .\) The mass of each block is $2.0 \mathrm{kg}$ and they are connected together by a cord. (a) If a horizontal force \(\overrightarrow{\mathbf{F}}\) pulling on block \(\mathrm{B}\) is slowly increased, in a direction parallel to the connecting cord, until it is barely enough to make the two blocks start moving, what is the magnitude of \(\overrightarrow{\mathbf{F}}\) at the instant that they start to slide? (b) What is the tension in the cord connecting blocks \(A\) and \(B\) at that same instant?
Four identical spring scales, \(\mathrm{A}, \mathrm{B}, \mathrm{C},\) and \(\mathrm{D}\) are used to hang a \(220.0-\mathrm{N}\) sack of potatoes. (a) Assume the scales have negligible weights and all four scales show the same reading. What is the reading of each scale? (b) Suppose that each scale has a weight of \(5.0 \mathrm{N} .\) If scales \(\mathrm{B}\) and \(\mathrm{D}\) show the same reading, what is the reading of each scale?
A helicopter of mass \(M\) is lowering a truck of mass \(m\) onto the deck of a ship. (a) At first, the helicopter and the truck move downward together (the length of the cable doesn't change). If their downward speed is decreasing at a rate of \(0.10 g,\) what is the tension in the cable? (b) As the truck gets close to the deck, the helicopter stops moving downward. While it hovers, it lets out the cable so that the truck is still moving downward. If the truck's downward speed is decreasing at a rate of \(0.10 g,\) while the helicopter is at rest, what is the tension in the cable?
A helicopter is lifting two crates simultaneously. One crate with a mass of \(200 \mathrm{kg}\) is attached to the helicopter by a cable. The second crate with a mass of \(100 \mathrm{kg}\) is hanging below the first crate and attached to the first crate by a cable. As the helicopter accelerates upward at a rate of \(1.0 \mathrm{m} / \mathrm{s}^{2},\) what is the tension in each of the two cables?
See all solutions

Recommended explanations on Physics Textbooks

Fluids

Read Explanation

Circular Motion and Gravitation

Read Explanation

Scientific Method Physics

Read Explanation

Magnetism

Read Explanation

Space Physics

Read Explanation

Nuclear 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.