M*g = 74.2 * 9.8 = 727 N. = Wt. of box = Normal force, Fn.
u*Fn = 727u = Force of kinetic friction.
Ft - u*Fn = M*a.
271 - 727u = 74.2*1.72.
u =
One day while moving boxes you get tired and decide to use a rocket instead. You attach a small rocket of negligible mass to a 74.2 kg box. When you turn the rocket on, it provides a constant thrust of 271 N, and the sbox begins sliding across the pavement. If the magnitude of acceleration of the box is 1.72 m/s2, what is the coefficient of kinetic friction between the soapbox and pavement?
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