|
|
|
|
|
| Author |
Message |
![[Post New]](/templates/default/images/icon_minipost_new.gif) 11 Apr 2008 19:25:45 IST
|
|
|
Pls Correct Me if I'm wrng...
The area under the force displacemnt graph gives the work done: Here it is 1/2 *10*6=30J. We know that the K.E. of the body at x=2 is 9J. Then the increase in k.e. frm that point onwards must be 30-9=21J.
Therefore 1/2 mv^2=21. On solving we get v=(21)1/2 =4.58 m s-1
|
this reply: 0 points
(with 0 
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
|
|
|
|
|
|
|
Moderation Team
![[Up]](/templates/default/images/icon_up.gif "[Up]"){kind=icon}
