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

Did you know? You can earn $20 for every friend you invite to Studypool!
Click here to
Refer a Friend
...
Oct 12th, 2015
...
Oct 12th, 2015
Feb 23rd, 2017
check_circle
Mark as Final Answer
check_circle
Unmark as Final Answer
check_circle
Final Answer

Secure Information

Content will be erased after question is completed.

check_circle
Final Answer