debra pushes the 25 kg box across the floor with a force of 134 newtons. find the magnitude of the acceleration of the box if the coeffiecient of kinetic friction between the box and the floor is .32 .
2007-01-19 13:13:35 · 3 answers · asked by judykharrison 1 in Science & Mathematics ➔ Physics
a=Fnet/m
since Friction and Fapply make up Fnet, we have:
a= Fapply-Friction / m
Now, you need to find Friction
Friction=mu(Fnormal)
Fnormal=Fweigh=mg=9.8*25=245N
Friction=.32(245N)=78.4N
a= 134N-78.4N / 25kg
a=2.22m/s^2
2007-01-19 13:23:00 · answer #1 · answered by 7 · 0⤊ 0⤋
m=25 kg
F=134 N
f=32 N
a=?
use the formula:
F=m*a
(F-f)=m*a
(134-32)=25*a
102=25a ---->>a=4.08
2007-01-19 21:28:04 · answer #2 · answered by frozentrain 2 · 0⤊ 0⤋
Easy. The answer is 4 m/s^2
2007-01-19 21:21:46 · answer #3 · answered by Anonymous · 0⤊ 0⤋