Question

find the work done in moving a particle along with a vector S=2i+5j-3k of the applied force F=5i-2i+3k

Answer 1

Answer:

HEY MATE

Explanation:

HERE'S IS YOUR ANSWER

AS WORK=FORCE×DISPLACEMENT IN DIRECTION OF FORCE

W=F. dS COS@

W=F VECTOR . dS vector

and according to dot product rules

multiple x direction component with x direction DISPLACEMENT an vice versa

W =FxDx + FyDy +FzDz

W=(5)(2)+(-2)(5)+(3)(-3)

W=10-10-9

so finally your answer came

W=-9 JOULE

W= -9 J

IT'S A NEGATIVE WORK

AND IT MEANS ON BODY IS LOSING/EMITTING ENERGY WHILE OTHER BODY UNABLE TO TAKE THAT ENERGY

HOPE IT WILL HELP YOU

THANKS