A force F = (7.10N)Î + (4.00 N)ĵ + (5.00 N) k acts on a 1.70 kg mobile object that moves from an initial position of di = (3.70 m) î + (8.10 m)ĵ + (2.00 m) k to a final position of df = (2.80 m)î + (8.50 m)} + (4.40 m) k in 1.70 s. Find (a) the work done on the object by the force in the 1.70 s interval, (b) the average power due to the force during that interval, and (c) the angle between vectors d; and df. (a) Number | Units (b) Number the tolerance is +/-5% |Units (c) Number Units

Physics for Scientists and Engineers, Technology Update (No access codes included)
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Chapter8: Conservation Of Energy
Section: Chapter Questions
Problem 8.62AP: A 1.00-kg object slides to the right on a surface having a coefficient of kinetic friction 0.250...
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Chapter 07, Problem 051
A force F = (7.10N)Î + (4.00 N)ĵ + (5.00 N) k acts on a 1.70 kg mobile object that
moves from an initial position of di = (3.70 m) î + (8.10 m)ĵ + (2.00 m) k to a final
position of df = (2.80 m)i + (8.50 m)j + (4.40 m) k in 1.70 s. Find (a) the work done
on the object by the force in the 1.70 s interval, (b) the average power due to the force
during that interval, and (c) the angle between vectors d; and df.
(a) Number
Units
(b) Number the tolerance is +/-5%
Units
(c) Number
Units
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Transcribed Image Text:Chapter 07, Problem 051 A force F = (7.10N)Î + (4.00 N)ĵ + (5.00 N) k acts on a 1.70 kg mobile object that moves from an initial position of di = (3.70 m) î + (8.10 m)ĵ + (2.00 m) k to a final position of df = (2.80 m)i + (8.50 m)j + (4.40 m) k in 1.70 s. Find (a) the work done on the object by the force in the 1.70 s interval, (b) the average power due to the force during that interval, and (c) the angle between vectors d; and df. (a) Number Units (b) Number the tolerance is +/-5% Units (c) Number Units Click if you would like to Show Work for this question: Open Show Work Question Attempts: 0 of 10 used SAVE FOR LATER SUBMIT ANSWER
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