PHYSICS EXERCISE

PHYSICS EXERCISE

BY:DIEGO GALLARDO

EXERCISE:

Given:Fg= 325 N Fp= 425N θ= 35.2° μk= 0.61 a= 0

A box of books weighing 325 N moves at a constant velocity across the floor when the box is pushed with a force of 425 N exerted downward at an angle of 35.2° below the horizontal. Find kinetic coefficient between the box and the floor.

Solution:

Fpy=Fp sin θ ∑Fy=Fpy+Fg+Fn=a =245+325+Fn=0 = 425N x sin (35.2) = 245N Fpx= (Fp cos θ ) Fn=245+325 =570 N =425 cos (35.2)= 347N

∑Fx=Fpx+Fk=a =347.29N + Fk = 0 Fk= 347

My kinetic coefficient is=0.61

Fk=ukFnuk=Fk/Fn

uk=347/570 =0.61