A force →F=(1.50N)ˆi+(1.10N)ˆj+(8.00N)ˆk acts on a 3.40 kg mobile object that moves from an initial position of →di=(4.20m)ˆi+(1.60m)ˆj+(5.10m)ˆk to a final position of →df=(2.80m)ˆi+(6.50m)ˆj+(2.80m)ˆk in 1.50 s. Find (a) the work done on the object by the force in the 1.50 s interval, (b) the average power due to the force during that interval, and (c) the angle between vectors →di and →df.
A force →F=(1.50N)ˆi+(1.10N)ˆj+(8.00N)ˆk acts on a 3.40 kg mobile object that moves from an initial position of →di=(4.20m)ˆi+(1.60m)ˆj+(5.10m)ˆk to a final position of →df=(2.80m)ˆi+(6.50m)ˆj+(2.80m)ˆk in 1.50 s. Find (a) the work done on the object by the force in the 1.50 s interval, (b) the average power due to the force during that interval, and (c) the angle between vectors →di and →df.
Principles of Physics: A Calculus-Based Text
5th Edition
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Raymond A. Serway, John W. Jewett
Chapter6: Energy Of A System
Section: Chapter Questions
Problem 12OQ
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A force →F=(1.50N)ˆi+(1.10N)ˆj+(8.00N)ˆk acts on a 3.40 kg mobile object that moves from an initial position of →di=(4.20m)ˆi+(1.60m)ˆj+(5.10m)ˆk to a final position of →df=(2.80m)ˆi+(6.50m)ˆj+(2.80m)ˆk in 1.50 s. Find (a) the work done on the object by the force in the 1.50 s interval, (b) the average power due to the force during that interval, and (c) the angle between vectors →di and →df.
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