a 100N block lies on a frictionless surface. A force of 50N was applied horizontally on the block moving it 5m. find the work done by the applied force and the done by the weight of the block
a 100N block lies on a frictionless surface. A force of 50N was applied horizontally on the block moving it 5m. find the work done by the applied force and the done by the weight of the block
University Physics Volume 1
18th Edition
ISBN:9781938168277
Author:William Moebs, Samuel J. Ling, Jeff Sanny
Publisher:William Moebs, Samuel J. Ling, Jeff Sanny
Chapter7: Work And Kinetic Energy
Section: Chapter Questions
Problem 101CP: The force F(x) varies with position, as shown beolow Find the work done by this force on a particle...
Related questions
Topic Video
Question
a 100N block lies on a frictionless surface. A force of 50N was applied horizontally on the block moving it 5m. find the work done by the applied force and the done by the weight of the block
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
This is a popular solution!
Trending now
This is a popular solution!
Step by step
Solved in 3 steps with 3 images
Follow-up Questions
Read through expert solutions to related follow-up questions below.
Follow-up Question
why 0(zero) at work done by the height of the block
![solution :-
The given situation is as follows
E = 50 N 2
Applied
farce.
:. The work
W₂
weight
F₂.dn'
& work done
is
done
7 dm² = 5m x
= 100N = 7₁ (Bay)
=
F
= 100 (-ŷ) N
applied force Fis
50 N X
5m x
= 50x5 N.m ‚Â
= 50X5 N.M
⇒ W₂ = 250 N.m = 250 J.
one by the weight force Fr of the bbck
W₁
F.dr
។ Lie
=
[4XX=1]
100N (-1). 5m x
=-100x5 (4.x) N.m
W₁ = 0
[+9₁8=0]](https://content.bartleby.com/qna-images/question/cf01c187-6139-4d96-a047-c30a7c0fa772/4233f4b5-bc11-40d0-9298-6623e3e31164/xgs088_processed.jpeg)
Transcribed Image Text:solution :-
The given situation is as follows
E = 50 N 2
Applied
farce.
:. The work
W₂
weight
F₂.dn'
& work done
is
done
7 dm² = 5m x
= 100N = 7₁ (Bay)
=
F
= 100 (-ŷ) N
applied force Fis
50 N X
5m x
= 50x5 N.m ‚Â
= 50X5 N.M
⇒ W₂ = 250 N.m = 250 J.
one by the weight force Fr of the bbck
W₁
F.dr
។ Lie
=
[4XX=1]
100N (-1). 5m x
=-100x5 (4.x) N.m
W₁ = 0
[+9₁8=0]
Solution
by Bartleby ExpertKnowledge Booster
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.Recommended textbooks for you
University Physics Volume 1
Physics
ISBN:
9781938168277
Author:
William Moebs, Samuel J. Ling, Jeff Sanny
Publisher:
OpenStax - Rice University
Glencoe Physics: Principles and Problems, Student…
Physics
ISBN:
9780078807213
Author:
Paul W. Zitzewitz
Publisher:
Glencoe/McGraw-Hill
An Introduction to Physical Science
Physics
ISBN:
9781305079137
Author:
James Shipman, Jerry D. Wilson, Charles A. Higgins, Omar Torres
Publisher:
Cengage Learning
University Physics Volume 1
Physics
ISBN:
9781938168277
Author:
William Moebs, Samuel J. Ling, Jeff Sanny
Publisher:
OpenStax - Rice University
Glencoe Physics: Principles and Problems, Student…
Physics
ISBN:
9780078807213
Author:
Paul W. Zitzewitz
Publisher:
Glencoe/McGraw-Hill
An Introduction to Physical Science
Physics
ISBN:
9781305079137
Author:
James Shipman, Jerry D. Wilson, Charles A. Higgins, Omar Torres
Publisher:
Cengage Learning
College Physics
Physics
ISBN:
9781938168000
Author:
Paul Peter Urone, Roger Hinrichs
Publisher:
OpenStax College
Classical Dynamics of Particles and Systems
Physics
ISBN:
9780534408961
Author:
Stephen T. Thornton, Jerry B. Marion
Publisher:
Cengage Learning
Physics for Scientists and Engineers, Technology …
Physics
ISBN:
9781305116399
Author:
Raymond A. Serway, John W. Jewett
Publisher:
Cengage Learning