A particle is acted upon by constant F1(2i^-3j^+4k^),F2(-i^+2k^-3k^)is displaced on the point A(2,1,0) to the point B(-3,-4,2)find the total work done?
Answers
Answered by
33
Hi , here is your answer !!
simple question !!!
A= ( 2,1,0)
B = ( -3 , -4 ,2)
B -A { why ? :- because we have to get change in displacement }
so , B - A => (-5 , -5 ,2 )
now , force => f2-f1
so , F = (-3i^ , 5k^ , 0 )
now , work done = F. S
so , FS = 15 -25 + 0
= -10 j
hence , net work done = is -10 j
hope it help you !!!
thanks !!
simple question !!!
A= ( 2,1,0)
B = ( -3 , -4 ,2)
B -A { why ? :- because we have to get change in displacement }
so , B - A => (-5 , -5 ,2 )
now , force => f2-f1
so , F = (-3i^ , 5k^ , 0 )
now , work done = F. S
so , FS = 15 -25 + 0
= -10 j
hence , net work done = is -10 j
hope it help you !!!
thanks !!
gautam7896101972:
Thank you
Answered by
12
Hi , here is your answer !!
simple question !!!
A= ( 2,1,0)
B = ( -3 , -4 ,2)
B -A { why ? :- because we have to get change in displacement }
so , B - A => (-5 , -5 ,2 )
now , force => f2-f1
so , F = (-3i^ , 5k^ , 0 )
now , work done = F. S
so , FS = 15 -25 + 0
= -10 j
hence , net work done = is -10 j
hope it help you !!!
thanks !!
simple question !!!
A= ( 2,1,0)
B = ( -3 , -4 ,2)
B -A { why ? :- because we have to get change in displacement }
so , B - A => (-5 , -5 ,2 )
now , force => f2-f1
so , F = (-3i^ , 5k^ , 0 )
now , work done = F. S
so , FS = 15 -25 + 0
= -10 j
hence , net work done = is -10 j
hope it help you !!!
thanks !!
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