Question

1.65 m/s 195 N 9.1 kg ? = 0.272 30° What is the magnitude of the work is done by the gravitational force? Answer in units of J. 004 (part 2 of 5) 10.0 points The work done by the gravitational field is 1. negative. 2. zero. 3. positive. 005 (part 3 of 5) 10.0 points How much work is done by the 195 N force? Answer in units of J. 006 (part 4 of 5) 10.0 points What is the change in kinetic energy of the crate? Answer in units of J. 007 (part 5 of 5) 10.0 points What is the speed of the crate after it is pulled 6.9 m? Answer in units of m/s. 

          1.65 m/s
195 N
9.1 kg
? = 0.272
30°
What is the magnitude of the work is done
by the gravitational force?
Answer in units of J.
004 (part 2 of 5) 10.0 points
The work done by the gravitational field is
1. negative.
2. zero.
3. positive.
005 (part 3 of 5) 10.0 points
How much work is done by the 195 N force?
Answer in units of J.
006 (part 4 of 5) 10.0 points
What is the change in kinetic energy of the
crate?
Answer in units of J.
007 (part 5 of 5) 10.0 points
What is the speed of the crate after it is pulled
6.9 m?
Answer in units of m/s.
Show more…
1.65 m/s 195 N 9.1 kg ? = 0.272 30° What is the magnitude of the work is done by the gravitational force? Answer in units of J. 004 (part 2 of 5) 10.0 points The work done by the gravitational field is 1. negative. 2. zero. 3. positive. 005 (part 3 of 5) 10.0 points How much work is done by the 195 N force? Answer in units of J. 006 (part 4 of 5) 10.0 points What is the change in kinetic energy of the crate? Answer in units of J. 007 (part 5 of 5) 10.0 points What is the speed of the crate after it is pulled 6.9 m? Answer in units of m/s.

Added by Laura G.

Close

University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
AceChat toggle button
Close icon
Ace pointing down

Please give Ace some feedback

Your feedback will help us improve your experience

Thumb up icon Thumb down icon
Thanks for your feedback!
Profile picture
A crate is pulled up a rough incline. The pulling force is parallel to the incline. The crate is pulled a distance of 6.9 m. The acceleration of gravity is 9.8 m/s 2. Is 195 N 1.65 m/ 9.1 kg : 0.272 30 What is the magnitude of the work is done by the gravitational force? Answer in units of J. 004 (part 2 of 5) 10.0 points The work done by the gravitational field is 1. negative. 2. zero. 3. positive. 005 (part 3 of 5) 10.0 points How much work is done by the 195 N force? Answer in units of J. 006 (part 4 of 5) 10.0 points What is the change in kinetic energy of the crate? Answer in units of J. 007 (part 5 of 5) 10.0 points What is the speed of the crate after it is pulled 6.9 m? Answer in units of m/s
Close icon
Play audio
Feedback
Powered by NumerAI
Kathleen Carty Danielle Fairburn
Ivan Kochetkov verified

Ivan Kochetkov and 57 other subject Physics 101 Mechanics educators are ready to help you.

Ask a new question
*

Labs

-

Want to see this concept in action?

NEW

Explore this concept interactively to see how it behaves as you change inputs.

View Labs
*

Key Concepts

-
Key Concept
Premium Feature
Explore the core concept behind this problem.
Play button
Key Concept
Premium Feature
Explore the core concept behind this problem.
Your browser does not support the video tag.
*

Recommended Videos

-
a-crate-of-mass-m-97-kg-is-pulled-up-rough-incline-with-an-initial-speed-of-vi-161-ms-the-pulling-force-is-f-105-n-parallel-to-the-incline-which-makes-an-angle-of-0-2030-with-the-horizontal-91463

Ivan K.
crate-of-mass-100-kg-pulled-up-rough-200-incline-the-pulling-force-100-n-parallel-to-the-incline-the-coefficient-of-kinetic-friction-is-04-and-the-crate-is-pulled-50m_-a-how-much-work-is-don-39743

Crate of mass 10.0 kg pulled up a rough 20° incline. The pulling force is 100 N parallel to the incline. The coefficient of kinetic friction is 0.4 and the crate is pulled 5.0 m. (a) How much work is done by gravity? (b) How much energy is lost due to friction? (c) How much work is done by the normal force? (d) What is the total or net work done on the crate? (e) If the package is pulled from rest, what is the speed of the crate after it has been pulled 5.0 m? Use the work-energy equation. (f) If the package is pulled with an initial speed of 1.5 m/s, what is the speed of the crate after it has been pulled 5.0 m? Use the work-energy equation.

Sri K.
a-crate-of-mass-94-kg-is-pulled-up-rough-incline-with-an-initial-speed-of-144-ms-the-pulling-force-is-102-n-parallel-to-the-incline-which-makes-an-angle-of-2030-with-the-horizontal-the-coeff-74726

A crate of mass 9.4 kg is pulled up a rough incline with an initial speed of 1.44 m/s. The pulling force is 102 N parallel to the incline, which makes an angle of 20.3° with the horizontal. The coefficient of kinetic friction is 0.400, and the crate is pulled 4.94 m. (a) How much work is done by the gravitational force on the crate? (b) Determine the increase in internal energy of the crate-incline system owing to friction. (c) How much work is done by the 102-N force on the crate? (d) What is the change in kinetic energy of the crate? (e) What is the speed of the crate after being pulled 4.94 m?

Ivan K.
*

Recommended Textbooks

-
University Physics with Modern Physics

University Physics with Modern Physics

Hugh D. Young 14th Edition
achievement 1,482 solutions
Physics: Principles with Applications

Physics: Principles with Applications

Douglas C. Giancoli 7th Edition
achievement 1,213 solutions
Fundamentals of Physics

Fundamentals of Physics

David Halliday, Robert Resnick , Jearl Walker 10th Edition
achievement 1,687 solutions
*

Transcript

-
00:01 Hello, here we have to first find the energy of work done by the gravitational force while the crate was moving.
00:10 So this work of the gravitational force equals to negative fgx times t.
00:20 And that is negative m g sine data times t.
01:04 So that's negative 162 jules.
01:09 Now we have to calculate increase in internal energy due to the friction and this increase equals to negative work of the friction force which is negative minus muqmg cosine tteta times t so that's 175 joules now in equation you have to calculate work done by the force f.
02:40 And this work equals to f times d...
Need help? Use Ace
Ace is your personal tutor. It breaks down any question with clear steps so you can learn.
Start Using Ace
Ace is your personal tutor for learning
Step-by-step explanations
Instant summaries
Summarize YouTube videos
Understand textbook images or PDFs
Study tools like quizzes and flashcards
Listen to your notes as a podcast
Continue solving this problem
Create a free account to:
  • View full step-by-step solution
  • Ask follow-up questions with Ace AI
  • Save progress and study later
Continue Free
Numerade

Get step-by-step video solution
from top educators

Continue with Clever
or



By creating an account, you agree to the Terms of Service and Privacy Policy
Already have an account? Log In

A free answer
just for you

Watch the video solution with this free unlock.

Numerade

Log in to watch this video
...and 100,000,000 more!


EMAIL

PASSWORD

OR
Continue with Clever