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
PHYSICS EXERCISE
dgallardo
Created on February 23, 2021
Start designing with a free template
Discover more than 1500 professional designs like these:
View
Genial Storytale Presentation
View
Historical Presentation
View
Scary Eighties Presentation
View
Psychedelic Presentation
View
Memories Presentation
View
Harmony Higher Education Thesis
View
Terrazzo Presentation
Explore all templates
Transcript
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