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sashaice [31]
3 years ago
7

A student pulls horizontally on a 12 kg box, which then moves horizontally with an acceleration of 0.2 m/s^2. If the student use

s a force of 15 N, what is the coefficient of kinetic friction of the floor?
Physics
1 answer:
polet [3.4K]3 years ago
7 0
The net force of the object is equal to the force applied minus the force of friction. 
                         Fnet = ma = F - Ff
                           12 kg x 0.2 m/s² = 15 N - Ff
The value of Ff is 12.6 N. This force is equal to the product of the normal force which is equal to the weight in horizontal surface and the coefficient of friction.
                             Ff = 12.6 N = k(12 kg)(9.81 m/s²)
The value of k is equal to 0.107. 
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I \omega_i = I \omega_f +mr^2 \omega_f\\\\mr^2 \omega_f = I \omega_i  - I \omega_f \\\\m = \frac{ I \omega_i  - I \omega_f}{r^2 \omega_f }

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