A 5.00-kg box slides 6.50 m across the floor before coming to rest. what is the coefficient of kinetic friction between the floor and the box if the box had an initial speed of 3.00 m/s?

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Physics

Jazlynn Hoover

8 November, 01:24

A 5.00-kg box slides 6.50 m across the floor before coming to rest. what is the coefficient of kinetic friction between the floor and the box if the box had an initial speed of 3.00 m/s?

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Gregory Strickland · Answer 1

Work=Fd and is equal to the change in energy, friction does work to stop the box. The box's kinetic energy is initially (K=.5m (v^2)) 22.5J and ends up at 0J. Friction does 22.5J of work over 6.5m. 22.5J = (F) (6.5m), F=3.46N. The force of friction is equal to the coefficient of friction times the normal force of an object, F=μFⁿ. We can determine the normal force by multiplying mass and gravity, Fⁿ=50N. Bringing this back to the force applied by friction, 3.46N/50N=μ=0.0692.