A force F~ = (6ˆi − 2 ˆj) N acts on a particle that undergoes a displacement ∆~r = (3ˆi + ˆj) m. Find (a) the work done by the force on the particle and (b) the angle between F~ and ∆~r.

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The magnitude of force is sqrt(6^ 2+2^2)= 6.3N with an angle of arctg-6/2=arctg(-3)=-71.56

The magnitude of displacement is sqrt(3^ 2+1^2)= 3.16m with an angle of arctg3/1=arctg(3)=71.56

The angle between F and delta r is 2*71.56=143.12

work= F. (delta r) cos teta= 6.3 *3.16*cos 143.12= -15.92 j

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