A 900-N crate is being pushed across a level floor at a constant speed by a forc

Question

A 900-N crate is being pushed across a level floor at a constant speed by a force of 250 N at an angle of 20.0° below the horizontal, as shown in the figure below.

(a) What is the coefficient of kinetic friction between the crate and the floor?
_______________________

(b) If the 250 N force is instead pulling the block at an angle of 20.0° above the horizontal as shown in the figure, what will be the acceleration of the crate? Assume that the coefficient of friction is the same as that found in part (a).____________ m/s2

Explanation / Answer

A) Assuming box is being pushed down the incline. constant speed so sum of all forces = 0 there for F applied + Fgrav = Ffriction Fgrav = 900sin20 Ffriction= 900sin20 + 250 Normal Force * coeff. fric. F norm = 900cos 20 so ...coeff. = 250 + 900sin20 / (900cos20) B) you have to find Fnet Fapplied -Fgravity-Ffriction=Fnet Ffriction=coeffic * 900cos20 Fgrav = 900sin20 Fapplied= 250 Mass =900/9.81 acc.= Fnet/mass Sorry no calculator

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