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.

Hello!

The force is a vector with the coordinates (6, -2) and the displacement (3, 1). Work is equal to the scalar product, i.e.

6*3 - 2*1 = 16. This is the answer for a).

For b), recall that tbe scalar product is cos of an angle if vectors are of length 1. Here we have to divide the scalar product by both lengths:

L1=sqrt(6^2+2^2)=sqrt(40)=2*sqrt(10),

L2=sqrt(3^2+1^2)=sqrt(10).

cos(a)=16/(2*sqrt(10)*sqrt(10))=16/20=0.8.

Therefore a=0.64rad.

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