PHYSICS
PHYSICS
5th Edition
ISBN: 2818440038631
Author: GIAMBATTISTA
Publisher: MCG
bartleby

Videos

Question
Book Icon
Chapter 6, Problem 39P

(a)

To determine

Speed of cart at point 3.

(a)

Expert Solution
Check Mark

Answer to Problem 39P

The speed is 14.3m/s.

Explanation of Solution

Speed of cart at point 1 is 20.0m/s and g is 9.81m/s2.

Potential energy is measured with respect to point 1.

Potential energy at the level of point 1 is taken as zero. Write the equation for total energy of at point 1.

E1=12mv12 (I)

Here, E1 is the total energy of cart at point 1, m is the mass of cart, and v1 is the velocity of cart at point 1.

Write the equation for total energy of at point 3.

E3=12mv32+mgy3 (II)

Here, E3 is the total energy of cart at point 3, v3 is the velocity of cart at point 3, g is the gravitational acceleration, and y3 is the height of point 3 from point 1.

According to the law of conservation of energy, the total energy at point 1 and 3 will remains the same. Equate the right hand sides equations (I) and (II).

12mv12=12mv32+mgy312v12=12v32+gy3

Rewrite the above equation in terms of v3.

v3=v122gy3 (III)

Conclusion:

Substitute 20.0m/s for v1, 9.81m/s2 for g, and 10.0m for y3 in the above equation to find v3.

v3=(20.0m/s)22(9.81m/s2)(10.0m)=204.49m/s=14.3m/s

Therefore, the speed is 14.3m/s.

(b)

To determine

Check whether the cart will reach the point 4 or not.

(b)

Expert Solution
Check Mark

Answer to Problem 39P

Yes. The car will reach point 4.

Explanation of Solution

Speed of cart at point 1 is 20.0m/s and g is 9.81m/s2.

Rewrite equation (III) by replacing v3 with v4.

v4=v122gy4

Here, v4 is the speed of cart at point 4 and y4 is the height of point 4 from point 1.

Write the condition for the cart to reach 4.

v12>2gy4

Conclusion:

Substitute 20.0m/s for v1, 9.81m/s2 for g, and 20.0m for y4 in the above equation to find whether cart reaches point 4or not.

(20.0m/s)2>2(9.81m/s2)(20.0m)400m2/s2>392.4m2/s2

Since the condition satisfies, cart can reach at point 4.

Substitute 20.0m/s for v1, 9.81m/s2 for g, and 20.0m for y4 in the equation for v4.

v4=(20.0m/s)22(9.81m/s2)(20.0m)=7.62m/s=2.76m/s

Therefore, the cart can reach at point 4.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
When current is flowing through the coil, the direction of the torque can be thought of in two ways. Either as the result of the forces on current carrying wires, or as a magnetic dipole moment trying to line up with an external field (e.g. like a compass). Note: the magnetic moment of a coil points in the direction of the coil's magnetic field at the center of the coil. d) Forces: We can consider the left-most piece of the loop (labeled ○) as a short segment of straight wire carrying current directly out of the page at us. Similarly, we can consider the right-most piece of the loop (labeled ) as a short segment straight wire carrying current directly into the page, away from us. Add to the picture below the two forces due to the external magnetic field acting on these two segments. Then describe how these two forces give a torque and determine if the torque acts to rotate the loop clockwise or counterclockwise according to this picture? B
In each of the following, solve the problem stated. Express your answers in three significant figures. No unit is considered incorrect. 1. For the circuit shown, determine all the currents in each branch using Kirchhoff's Laws. (3 points) 6 5V 2 B C 4 A www 6 VT ww T10 V F E 2. Compute for the total power dissipation of the circuit in previous item. (1 point) 3. Use Maxwell's Mesh to find Ix and VAB for the circuit shown. (3 points) Ix 50 V 20 ww 21x B 4. Calculate all the currents in each branch using Maxwell's Mesh for the circuit shown. (3 points) www 5ი 10 24V 2A 2002 36V
If the mass of substance (1 kg), initial temperature (125˚C), the final temperature (175˚C) and the total volume of a closed container (1 m3) remains constant in two experiments, but one experiment is done with water ( ) and the other is done with nitrogen ( ). What is the difference in the change in pressure between water and nitrogen?

Chapter 6 Solutions

PHYSICS

Ch. 6.4 - Prob. 6.6PPCh. 6.5 - Prob. 6.5CPCh. 6.5 - Prob. 6.7PPCh. 6.5 - Prob. 6.8PPCh. 6.6 - Prob. 6.9PPCh. 6.6 - Prob. 6.10PPCh. 6.7 - Prob. 6.7CPCh. 6.7 - Practice Problem 6.11 A Misfire The same dart gun...Ch. 6.7 - Prob. 6.12PPCh. 6.8 - Prob. 6.13PPCh. 6.8 - Prob. 6.14PPCh. 6 - Prob. 1CQCh. 6 - Prob. 2CQCh. 6 - Prob. 3CQCh. 6 - Prob. 4CQCh. 6 - Prob. 5CQCh. 6 - Prob. 6CQCh. 6 - Prob. 7CQCh. 6 - Prob. 8CQCh. 6 - Prob. 9CQCh. 6 - Prob. 10CQCh. 6 - Prob. 11CQCh. 6 - Prob. 12CQCh. 6 - Prob. 13CQCh. 6 - Prob. 1MCQCh. 6 - 2. If a kangaroo on Earth can jump from a standing...Ch. 6 - Prob. 3MCQCh. 6 - Prob. 4MCQCh. 6 - Prob. 5MCQCh. 6 - Prob. 6MCQCh. 6 - Prob. 7MCQCh. 6 - Prob. 8MCQCh. 6 - Prob. 9MCQCh. 6 - Questions 9 and 10. A simple catapult, consisting...Ch. 6 - Prob. 11MCQCh. 6 - Prob. 1PCh. 6 - Prob. 2PCh. 6 - Prob. 3PCh. 6 - Prob. 4PCh. 6 - Prob. 5PCh. 6 - Prob. 6PCh. 6 - Prob. 7PCh. 6 - Prob. 8PCh. 6 - Prob. 9PCh. 6 - Prob. 10PCh. 6 - Prob. 11PCh. 6 - Prob. 12PCh. 6 - Prob. 13PCh. 6 - Prob. 14PCh. 6 - Prob. 15PCh. 6 - Prob. 16PCh. 6 - Prob. 17PCh. 6 - Prob. 18PCh. 6 - Prob. 19PCh. 6 - Prob. 20PCh. 6 - Prob. 21PCh. 6 - Prob. 22PCh. 6 - Prob. 23PCh. 6 - Prob. 24PCh. 6 - Prob. 25PCh. 6 - Prob. 26PCh. 6 - Prob. 27PCh. 6 - Prob. 28PCh. 6 - Problems 29–32. A skier passes through points A–E...Ch. 6 - Prob. 30PCh. 6 - Prob. 31PCh. 6 - Prob. 32PCh. 6 - Prob. 33PCh. 6 - Prob. 34PCh. 6 - 35. Emil is tossing an orange of mass 0.30 kg into...Ch. 6 - Prob. 36PCh. 6 - 37. An arrangement of two pulleys, as shown in the...Ch. 6 - Prob. 38PCh. 6 - Prob. 39PCh. 6 - Prob. 40PCh. 6 - Prob. 41PCh. 6 - Prob. 42PCh. 6 - Prob. 43PCh. 6 - Prob. 44PCh. 6 - Prob. 45PCh. 6 - Prob. 46PCh. 6 - 47. Refer to Problems 11-14. Find conservation of...Ch. 6 - Prob. 48PCh. 6 - Prob. 49PCh. 6 - Prob. 50PCh. 6 - Prob. 51PCh. 6 - Prob. 52PCh. 6 - 53. What is the minimum speed with which a meteor...Ch. 6 - 54. A projectile with mass of 500 kg is launched...Ch. 6 - Prob. 55PCh. 6 - Prob. 56PCh. 6 - Prob. 57PCh. 6 - Prob. 58PCh. 6 - Prob. 59PCh. 6 - Prob. 60PCh. 6 - Prob. 61PCh. 6 - Prob. 62PCh. 6 - Prob. 63PCh. 6 - Prob. 64PCh. 6 - Prob. 65PCh. 6 - Prob. 66PCh. 6 - Prob. 67PCh. 6 - Prob. 68PCh. 6 - Prob. 69PCh. 6 - Prob. 70PCh. 6 - Prob. 71PCh. 6 - Prob. 72PCh. 6 - Prob. 73PCh. 6 - Prob. 74PCh. 6 - Prob. 75PCh. 6 - Prob. 76PCh. 6 - Prob. 77PCh. 6 - Prob. 78PCh. 6 - Prob. 79PCh. 6 - Prob. 80PCh. 6 - Prob. 81PCh. 6 - Prob. 82PCh. 6 - Prob. 83PCh. 6 - Prob. 84PCh. 6 - Prob. 85PCh. 6 - Prob. 86PCh. 6 - Prob. 87PCh. 6 - Prob. 88PCh. 6 - Prob. 89PCh. 6 - Prob. 90PCh. 6 - Prob. 91PCh. 6 - Prob. 92PCh. 6 - Prob. 93PCh. 6 - Prob. 94PCh. 6 - Prob. 95PCh. 6 - Prob. 97PCh. 6 - Prob. 96PCh. 6 - Prob. 98PCh. 6 - Prob. 99PCh. 6 - Prob. 101PCh. 6 - Prob. 100PCh. 6 - Prob. 102PCh. 6 - Prob. 103PCh. 6 - Prob. 104PCh. 6 - Prob. 105PCh. 6 - Prob. 107PCh. 6 - Prob. 108PCh. 6 - Prob. 109PCh. 6 - Prob. 110PCh. 6 - Prob. 111PCh. 6 - Prob. 112PCh. 6 - Prob. 106PCh. 6 - Prob. 113PCh. 6 - Prob. 114PCh. 6 - Prob. 115PCh. 6 - Prob. 116PCh. 6 - Prob. 117PCh. 6 - Prob. 118PCh. 6 - Prob. 119PCh. 6 - Prob. 120PCh. 6 - Prob. 130PCh. 6 - Prob. 125PCh. 6 - Problems 121 and 122.A particle is constrained to...Ch. 6 - Prob. 122PCh. 6 - Prob. 123PCh. 6 - Prob. 124PCh. 6 - Prob. 132PCh. 6 - Prob. 126PCh. 6 - Prob. 127PCh. 6 - Prob. 128PCh. 6 - Prob. 129PCh. 6 - Prob. 131PCh. 6 - Prob. 142PCh. 6 - Prob. 140PCh. 6 - Prob. 133PCh. 6 - Prob. 134PCh. 6 - Prob. 136PCh. 6 - Prob. 137PCh. 6 - Prob. 138PCh. 6 - Prob. 135PCh. 6 - Prob. 139PCh. 6 - Prob. 141P
Knowledge Booster
Background pattern image
Physics
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
College Physics
Physics
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Cengage Learning
Text book image
University Physics (14th Edition)
Physics
ISBN:9780133969290
Author:Hugh D. Young, Roger A. Freedman
Publisher:PEARSON
Text book image
Introduction To Quantum Mechanics
Physics
ISBN:9781107189638
Author:Griffiths, David J., Schroeter, Darrell F.
Publisher:Cambridge University Press
Text book image
Physics for Scientists and Engineers
Physics
ISBN:9781337553278
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Text book image
Lecture- Tutorials for Introductory Astronomy
Physics
ISBN:9780321820464
Author:Edward E. Prather, Tim P. Slater, Jeff P. Adams, Gina Brissenden
Publisher:Addison-Wesley
Text book image
College Physics: A Strategic Approach (4th Editio...
Physics
ISBN:9780134609034
Author:Randall D. Knight (Professor Emeritus), Brian Jones, Stuart Field
Publisher:PEARSON
Mechanical work done (GCSE Physics); Author: Dr de Bruin's Classroom;https://www.youtube.com/watch?v=OapgRhYDMvw;License: Standard YouTube License, CC-BY