homework help > physics > physics q&a board > topic: work...

Q BgQuestion:

      
Novice
Karma Points: 25
Respect (79%):
posted by  dezzzzz on 11/6/2009 2:46:27 PM  |  status: Live  |  Earned Karma: 25

work

Course Textbook Chapter Problem Needs by
N/A N/A N/A N/A 11/9/2009 at 12:00:00 PM
Question Details:

A cutting tool under microprocessor control has several forces acting on it. One force is \vec{F} = - \alpha xy^{2} \hat{ j }, a force in the negative y-direction whose magnitude depends on the position of the tool. The constant is alpha = 2.40 N/m. Consider the displacement of the tool from the origin to the point x = 3.40 m, y = 3.40 m.

Calculate the work done on the tool by \vec{F} if this displacement is along the straight line y=x that connects these two points.
Calculate the work done on the tool by \vec{F} if the tool is first moved out along the x-axis to the point x = 3.40 m, y=0 and then moved parallel to the y-axis to x= 3.40 m, y= 3.40 m.
Tags: Physics
Bonus Point Alert! Earn +4 additional karma points for helping this gold member.

AAnswers:

Answer Question Ask for clarification
(SME)
posted by christina.S on 11/7/2009 9:41:21 AM  |  status: Live
Asker's Rating: Helpful   
Response Details:
Given :
.
               
.
               Let displacement is :  S =  x i  +  y j
.
(a)  We know that :
.
               Work done (W)  =  F .ΔS   =  ( - α xy2 j ). (x i  +  y j ) 
.
                                                       =   - α xy3    =   - 2.40 * (3.40 ) * ( 3.40)3    =  --------  J
.
(b)   Here  ΔS  =  ( 3.40  i  +  0 j )  -  (3.40 i   +  3.40 j )   =  - 3.40  j
.
       Hence work done  is :
.
                                 W  =   F .ΔS   =   ( - α xy2 j ). (- 3.40 j) 
.
                                                        =    - 2.40 * (3.40 )2* ( - 3.40)   =   --------- J
.
Solve the above
.
Hope this helps u!  
Answer Question Ask for clarificarion
Ask New Question

Join Cramster's Community

Cramster.com brings together students, educators and subject enthusiasts in an online study community. With around-the-clock expert help and a community of over 100,000 knowledgeable members, you can find the help you need, whenever you need it. Join for free today » How Cramster is different from tutoring »

Need Help In

Useful Links