A 64 kg box is resting on a carpet floor. If thecoefficient of static friction b

Question

A 64 kg box is resting on a carpet floor. If thecoefficient of static friction between the box and floor is 0.98... a) What force is required to put the box in motion? b) Once the box reaches a speed of 2.5 m/s, what net forcewill be requierd to keep the box moving att his speed for 30s?Explain. Btw, how do you figure out the coefficient of kineticfriction? :S Thank you! A 64 kg box is resting on a carpet floor. If thecoefficient of static friction between the box and floor is 0.98... a) What force is required to put the box in motion? b) Once the box reaches a speed of 2.5 m/s, what net forcewill be requierd to keep the box moving att his speed for 30s?Explain. Btw, how do you figure out the coefficient of kineticfriction? :S Thank you!

Explanation / Answer

Given : Mass ( m ) = 64 kg Coefficient of static friction ( s ) = 0.98 g = 9.8 m/s2 Force ( F ) = ? We know that   s =Fs / N Where   N = m g    Fs = s N         =s m g        = ------- N Solve the remaining I hope it helps you

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