EBK PHYSICS FOR SCIENTISTS AND ENGINEER
EBK PHYSICS FOR SCIENTISTS AND ENGINEER
6th Edition
ISBN: 9781319321710
Author: Mosca
Publisher: VST
bartleby

Videos

Question
Book Icon
Chapter 6, Problem 28P

(a)

To determine

The work done by the force from x=0.0 to x=2.0m .

(a)

Expert Solution
Check Mark

Explanation of Solution

Given:

Mass of the object is 3.0kg .

Velocity of the object is 2.4m/s .

Formula used:

Write expression for work done by the force from x=0.0 to x=2.0m .

  W=nA  ........ (1)

Here, n is number of squares and A is area of one square.

Write expression for area of square.

  A=xy  ........ (2)

Here, x is value of distance covered in one unit and y is value of force applied.

Calculation:

Substitute 0.25m for x and 0.5N for y in equation (2).

  A=(0.25m)(0.5N)A=0.125J

Substitute 22 for n and 0.125J for A in equation (1).

  W=(22)(0.125J)W=2.75J

Conclusion:

Thus, work done by the force is 2.75J .

(b)

To determine

The kinetic energy of object at x=2.0m .

(b)

Expert Solution
Check Mark

Explanation of Solution

Given:

Mass of the object is 3.0kg .

Velocity of the object is 2.4m/s .

Formula used:

Write expression for work done by the force from x=0.0 to x=2.0m .

  W=nA  ........ (1)

Here, n is number of squares and A is area of one square.

Write expression for area of square.

  A=xy  ........ (2)

Here, x is value of distance covered in one unit and y is value of force applied.

Write expression for kinetic energy at x=2.0m .

  K2m=K0+W

Substitute 12mv2 for K0 in above expression.

  K2m=12mv2+W  ........ (3)

Calculation:

Substitute 0.25m for x and 0.5N for y in equation (2).

  A=(0.25m)(0.5N)A=0.125J

Substitute 22 for n and 0.125J for A in equation (1).

  W=(22)(0.125J)W=2.75J

Substitute 3.0kg for m , 2.4m/s2 and 2.75J in equation (3).

  K2m=12(3.0kg)(2.4m/s2)2+2.75JK2m=11.4J

Conclusion:

Thus, kinetic energy at x=2.0m is 11.4J .

(c)

To determine

The speed of object at x=2.0m .

(c)

Expert Solution
Check Mark

Explanation of Solution

Given:

Mass of the object is 3.0kg .

Velocity of the object is 2.4m/s .

Formula used:

  K2m=12mv2+W

Write expression for velocity at x=2.0m .

  v2m=2K2mm

Calculation:

Substitute 11.4J for K2m and 3.0kg for m

  v2m=2(11.4J)3.0kgv2m=2.8m/s

Conclusion:

Thus, the velocity of object is 2.8m/s .

(d)

To determine

The work done on the object from x=0.0m to x=4.0m .

(d)

Expert Solution
Check Mark

Explanation of Solution

Given:

Mass of the object is 3.0kg .

Velocity of the object is 2.4m/s .

Formula used:

Write expression for work done on the object from x=0.0m to x=4.0m .

  W=nA  ........ (1)

Here, n is number of squares and A is area of one square.

Write expression for area of square.

  A=xy  ........ (2)

Here, x is value of distance covered in one unit and y is value of force applied.

Calculation:

Substitute 0.25m for x and 0.5N for y in equation (2).

  A=(0.25m)(0.5N)A=0.125J

Substitute 26 for n and 0.125J for A in equation (1).

  W=(26)(0.125J)W=3.25J

Conclusion:

Thus, work done is 3.25J .

(e)

To determine

The speed of object at x=4.0m .

(e)

Expert Solution
Check Mark

Explanation of Solution

Given:

Mass of the object is 3.0kg .

Velocity of the object is 2.4m/s .

Formula used:

Write expression for kinetic energy at x=4.0m .

  K4m=K0+W

Substitute 12mv4m2 for K0 in above expression.

  K4m=12mv4m2+W  ........ (1)

Write expression for velocity at x=4.0m .

  v4m=2K4mm  ........ (2)

Calculation:

Substitute 3.0kg for m , 2.4m/s2 and 3.25J in equation (3).

  K4m=12(3.0kg)(2.4m/s2)2+3.25JK4m=11.9J

Substitute 11.4J for K2m and 3.0kg for m in equation (4).

  v4m=2(11.9J)3.0kgv4m=2.8m/s

Conclusion:

Thus, velocity of the object is 2.8m/s .

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
suggest a reason ultrasound cleaning is better than cleaning by hand?
Checkpoint 4 The figure shows four orientations of an electric di- pole in an external electric field. Rank the orienta- tions according to (a) the magnitude of the torque on the dipole and (b) the potential energy of the di- pole, greatest first. (1) (2) E (4)
What is integrated science. What is fractional distillation What is simple distillation
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
Physics for Scientists and Engineers: Foundations...
Physics
ISBN:9781133939146
Author:Katz, Debora M.
Publisher:Cengage Learning
Text book image
University Physics Volume 1
Physics
ISBN:9781938168277
Author:William Moebs, Samuel J. Ling, Jeff Sanny
Publisher:OpenStax - Rice University
Text book image
Principles of Physics: A Calculus-Based Text
Physics
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Text book image
Glencoe Physics: Principles and Problems, Student...
Physics
ISBN:9780078807213
Author:Paul W. Zitzewitz
Publisher:Glencoe/McGraw-Hill
Text book image
Physics for Scientists and Engineers, Technology ...
Physics
ISBN:9781305116399
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Text book image
College Physics
Physics
ISBN:9781938168000
Author:Paul Peter Urone, Roger Hinrichs
Publisher:OpenStax College
Mechanical work done (GCSE Physics); Author: Dr de Bruin's Classroom;https://www.youtube.com/watch?v=OapgRhYDMvw;License: Standard YouTube License, CC-BY