A force É = (1.90 N )î + (1.00 N )ŷ + (4.30 N )k acts on a 5.70 kg mobile object that moves from an initial position of d i = (5.30 m )î + (6.30 m )ĵ + (2.70 m )k to a final position of á f = (8.90 m )î + (3.60 m ŷ + (5.40 m )k in 4.70 s. Find (a) the work done on the object by the force in the 4.70 s interval, (b) the average power due to the force during that m j + interval, and (c) the angle between vectors d i and á f. (a) Number i Units (b) Number i Units (c) Number i Units

Physics for Scientists and Engineers with Modern Physics
10th Edition
ISBN:9781337553292
Author:Raymond A. Serway, John W. Jewett
Publisher:Raymond A. Serway, John W. Jewett
Chapter7: Energy Of A System
Section: Chapter Questions
Problem 19P: (a) A force F=(4xi+3yj), where F is in newtons and x and y are in meters, acts on an object as the...
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A force F
= (1.90 N )i + (1.00 N )j + (4.30 N )k acts on a 5.70 kg mobile object that moves from an initial position of
(8.90 m )i + (3.60 m )j + (5.40 m )k in
4.70 s. Find (a) the work done on the object by the force in the 4.70 s interval, (b) the average power due to the force during that
i =
(5.30 m )i + (6.30 m )j + (2.70 m )k to a final position of d
f =
interval, and (c) the angle between vectors d ¡ and d
f.
(a) Number
i
Units
(b) Number
Units
(c) Number
i
Units
Transcribed Image Text:A force F = (1.90 N )i + (1.00 N )j + (4.30 N )k acts on a 5.70 kg mobile object that moves from an initial position of (8.90 m )i + (3.60 m )j + (5.40 m )k in 4.70 s. Find (a) the work done on the object by the force in the 4.70 s interval, (b) the average power due to the force during that i = (5.30 m )i + (6.30 m )j + (2.70 m )k to a final position of d f = interval, and (c) the angle between vectors d ¡ and d f. (a) Number i Units (b) Number Units (c) Number i Units
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