12 replies to this topic

[Rush]

If you don't know anything about physics, don't spam my thread stating you don't know or good at physics, mmkay?
Okay the book I have is:
Physics Principles with Apllications
4th Edition
Douglas C. Giancoli
Prentice Hall International, Inc.
Englewood Cliffs, New Jersey 07632

The PROBLEM ( not question) in the book is on:
Section 6-1 (about work, force, etc.)
#2a and #2b

2. (I) A 900-N crate rests on the floor. How much work is requred to move it at constant speed (a) 4.0m along the floor against a friction force of 180N, and
(b) 4.0m vertically

I need help solving it.

It doesn't state the applied force..
I know the equation work=(force)(distance)(cos theta/angle)
Also the one for ramps..

Edited by Rush, 08 November 2006 - 05:41 AM.

[Sweeney]

At a constant speed, the applied force is the same as the frictional force.
Otherwise the crate would be accelerating.

Can you solve it now?

Note to everyone else, and I've said this before, please do not solve the problem for him. By all means, offer help and advice, but noone learns from being given an answer.

[Gnome]

Good luck with finding the answer!

[Rush]

Thanks Joe, im solving it in a jiffy using an applied force of 180N with a frictional force of 180N to keep it in constant motion with a distance of 4m..and applied at ang angle of zero..Thanks again
^from what I understood

[Sweeney]

Sounds about right.
The same rule applies for the vertical one too, only this time you have a weight opposing the motion instead of a frictional force.

Post the answers you get, and I'll check them, if you like

[Rush]

(a) Work=(180 N)(4.0 meters)(cos0)
=720 Joules
I hope i got this right.
cos0=1 so it doesn't affect the equation..
180 N was used to negate to frictional force
4.0 meters was the distance required

(b) Work= (900 N)(4.0 meters)(-see note below-)
= 3600 J
If I made it cos90, it would become zero
so I just left it out..

So there

[Plew]

man i havent even learned this in my physics class i am still on velocity equations

Edited by Plew, 08 November 2006 - 06:31 AM.

[Rush]

What year are you? Just curious, not assuming

Edited by Rush, 08 November 2006 - 06:33 AM.

[Sweeney]

The rule for slopes only applies if the object is pulled up a rough slope.
A vertical rise does not require it

They look right to me, good job

[Bryan]

Oh, wait until you get into torque. Fun stuff right there.

[Redblade]

Oh, wait until you get into torque. Fun stuff right there.

hmm its called moments here. Mechanics are one of my best topics. anyway for the second answer, the angle is still 0, not 90 and it is with reference to the direction of the force you are applying to, so there is some conceptual mistake. It would be 90 in a scenario eg, A man walking across a room carrying a bag of groceries. Since direction of force was upwards but direction of distance moved say to is right the angle between would be 90 degree and therefore no work is done. ( thats the best i could explain without diagrams )

P.S i am into quantum mechanics now. you could pm me if you're studying it and had any problems.

[Shortieking]

Good luck finding the answer.
I'll be taking Physics next year.
Wish I could have helped =/

[Sweeney]

hmm its called moments here. Mechanics are one of my best topics. anyway for the second answer, the angle is still 0, not 90 and it is with reference to the direction of the force you are applying to, so there is some conceptual mistake. It would be 90 in a scenario eg, A man walking across a room carrying a bag of groceries. Since direction of force was upwards but direction of distance moved say to is right the angle between would be 90 degree and therefore no work is done. ( thats the best i could explain without diagrams )

P.S i am into quantum mechanics now. you could pm me if you're studying it and had any problems.

That's assuming the bag handles stayed totally vertical, which would never happen in practise, of course.
A good example though.