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

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

University Physics Volume 1

18th Edition

ISBN:9781938168277

Author:William Moebs, Samuel J. Ling, Jeff Sanny

Publisher:William Moebs, Samuel J. Ling, Jeff Sanny

Chapter7: Work And Kinetic Energy

Section: Chapter Questions

Problem 88AP: Consider a particle on which a force acts that depends on the position of the particle. This force...

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