Physics problem to solve

Physics
Tutor: None Selected Time limit: 1 Day

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.

Oct 12th, 2015

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.

Please ask if something is unclear.
Oct 12th, 2015

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Oct 12th, 2015
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