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Three constant forces F1 = 2i-3j+2k, F2 = i+j-k, and F3 = 3i+j-2k in newtons displace a particle from (1, -1, 2) to (-1, -1, 3) and then to (2, 2, 0) (displacement being measured in meteres). Find the total work done by the forces.

shubham , 6 Years ago
Grade 12
anser 2 Answers
HIMANSHU SHEKHAR DAS

Last Activity: 6 Years ago

we know that work done W=F.S, here both F and S are vectors, and dot represents dot product.
where F is net force and S is  net displacement as work done is a conservative force.
net force acting on the body F=F1+F2+F3
so net force comes out to be 6i-j-k
and net displacement is from (1,-1,2) to (2,2,0) which is 2i+2j+0k-1i-1j+2k= i+j-2k
so total work done is (6i-j-k).(i+j-2k)= 7 J
so total work done is 7 J
thanks

YHB

Last Activity: 4 Years ago

f1= 2i-3j+2k
f2=i+k-k
f3=3i+j*-2k
add each terms separately(ie i cap terms separately,j cap terms separately, k cap terms separately)
i cap=2+3+1= 6 i
j cap terms=-3+1+1=-j
k cap terms=2-1-2=-k
6i-j-k-------(1)
substract (1,-1,2) from(-1,-1,3) and mark it as s1 =-2i+0j+l----(2)
substract (-1,-1,3) from (2,2,0) and mark it as s2=3i+3j-3k-----(3)
 
case1
-----------
multiply (1) and (2)
-12+0-1=-13j----------(4)
 
case(2)
----------- 
 
multiply (1) and (3)
18-3+3=18--------(5)
 
now add equations (4) and (5) to get the total work done by the forces
 
total work done=18-13=5J
 
HENCE THE FINAL ANSWER IS 5J FOR THIS QUESTION 
THANK YOU!!!!
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