A constant applied force F, of 15.0 N pushes a box with a mass m = 7.00 kg a distance x = 15.0 m across a level floor. The coefficient of kinetic friction between the box and the floor is 0.150. Fp m Assuming that the box starts from rest, what is the final velocity vf of the box at the 15.0 m point? Uf m/s If there were no friction between the box and the floor, what applied force Fnew would give the box the same final velocity? Fnew = N

Physics for Scientists and Engineers, Technology Update (No access codes included)
9th Edition
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Author:Raymond A. Serway, John W. Jewett
Publisher:Raymond A. Serway, John W. Jewett
Chapter5: The Laws Of Motion
Section: Chapter Questions
Problem 5.61P: Review. A 3.00-kg block starts from rest at the top of a 30.0 incline and slides a distance of 2.00...
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A constant applied force ? of 15.0 N pushes a box with a mass ?=7.00 kg a distance ?=15.0 m across a level floor. The coefficient of kinetic friction between the box and the floor is 0.150.

A constant applied force F, of 15.0 N pushes a box with a mass m = 7.00 kg a distance x = 15.0 m across a level floor.
The coefficient of kinetic friction between the box and the floor is 0.150.
Fp
m
Assuming that the box starts from rest, what is the final velocity vf of the box at the 15.0 m point?
Uf =
m/s
If there were no friction between the box and the floor, what applied force Fnew would give the box the same final velocity?
Fnew =
N
Transcribed Image Text:A constant applied force F, of 15.0 N pushes a box with a mass m = 7.00 kg a distance x = 15.0 m across a level floor. The coefficient of kinetic friction between the box and the floor is 0.150. Fp m Assuming that the box starts from rest, what is the final velocity vf of the box at the 15.0 m point? Uf = m/s If there were no friction between the box and the floor, what applied force Fnew would give the box the same final velocity? Fnew = N
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