Question involving LAWS OF WORK

[demode]

I have been given the prompt:

"Using a set of pulleys, a mover raises a 1000-N crate a distance of 5.00 m. This is accomplished by pulling in 20.0 m of cord"

I need to find

• How much effort would have to be applied if this were an ideal machine
• If the effort actually applied is 300 N, what force is used to overcome friction?
• What is the work output?
• What is the work input?
• What is the real mechanical advantage?

I don't know if I have done this correctly so if someone would be willing to verify my answers or guide me through this, it would be appreciated.


1. In an ideal machine, I believe work input is equal to work output. With that being said, I can set up the equation 1000N * 5.0 M = XN * 20.0 M, then I can solve X for 250 N... Correct?

2. I'm not too sure what it means when it asks to OVERCOME friction, so what do i do?

3. Work output is just equal to force * distance, so we evaluate the given 1000N force times the given 5.0 M distance and get 5000 J, Correct?

4. Work input is just equal to force * distance as well, so we evaluate the value we solved for in number 1 (250) and multiply by the given 25 m distance and get 5000J, correct?

5. Not sure at all what to do

 

[PhanthomJay]

I have been given the prompt:

"Using a set of pulleys, a mover raises a 1000-N crate a distance of 5.00 m. This is accomplished by pulling in 20.0 m of cord"

I need to find

• How much effort would have to be applied if this were an ideal machine
• If the effort actually applied is 300 N, what force is used to overcome friction?
• What is the work output?
• What is the work input?
• What is the real mechanical advantage?

I don't know if I have done this correctly so if someone would be willing to verify my answers or guide me through this, it would be appreciated.


1. In an ideal machine, I believe work input is equal to work output. With that being said, I can set up the equation 1000N * 5.0 M = XN * 20.0 M, then I can solve X for 250 N... Correct?yes[/color]
2. I'm not too sure what it means when it asks to OVERCOME friction, so what do i do?Idealy it would take 250N of force, but it took 300N. So why was the extra 50N required?[/color]

3. Work output is just equal to force * distance, so we evaluate the given 1000N force times the given 5.0 M distance and get 5000 J, Correct?yes[/color]

4. Work input is just equal to force * distance as well, so we evaluate the value we solved for in number 1 (250) the problem wants you to use 300N here, not the ideal 250N[/color] and multiply by the given 25 m you mean 20m[/color]distance and get 5000J, correct?that works out to 6000J[/color]

5. Not sure at all what to do

5. Ideal MA is 4, since it took just 1/4 of the weight to move it (250/1000 =1/4). So its actual MA is??[/color]

 

[demode]

alright man, I understand it. Thanks SO much