The answer for the following question is posted below please show how you got th

Question

The answer for the following question is posted below please show how you got the answer.

9. A force F= 3.00 i 7.00 j +7.00 k (in N) acts on a 2.00 kg object that moves from an initial position of 3.00 7-2.00 + 5.00 k (in m) to a final position =5.00 + 4.00+ 7.00k (also in m) in a time a) the work done on the object by the force in the 4.00 s interval, b) the average power due to the force during that same interval, and c) the angle between vectors di and d d. Answers: a) 32.0 J b) 8.00 w c) 78.2

Explanation / Answer

a) Force applied = 3i+7j+7k

initial position = 3i-2j+5k

Final position=-5i+4j+7k

Displacement = final position-initial position = -8i+6j+2k

Work done is given by F.d = (3i+7j+7k).(-8i+6j+2k)=32 joules

b)Time taken =4 seconds

Average power=Work/time=32/4=8 Watts

c)magnitude of di=sqrt(32+22+52)=sqrt(38)

magniutude of df = sqrt(52+42+72)=sqrt(90)

dot product of di and df = 12

angle between them = cos-1(12/sqrt(38)*sqrt(90))=78.2o

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