a 412 kg crate is at rest on a level surface. at t = 0 seconds, a horizontal force of 1630 newtons is applied to the crate to give it a constant acceleration to the left. (a) given that the coefficient of kinetic friction is .25, what is the amount of kinetic friction force? (b) use the impulse-momentum equation to find the velocity of the crate at t = 3.5 secs.
2007-02-20 14:46:46 · 2 answers · asked by mandy 1 in Science & Mathematics ➔ Physics
a)friction is fun f=un
f=412(9.8)(.25)
f=1009.4N
b)1630-1009.4=620.6N
F=ma
620.6=412a
a=1.51m/s
vf=vi+at
vf=0+(1.51)(3.5)
vf=5.27m/s
2007-02-20 14:55:07 · answer #1 · answered by climberguy12 7 · 0⤊ 0⤋
Force friction = mu x forcenormal = .25 x 412 x 9.8 N
Force net = force applied - force friction
Impulse = Force net x time = mass x change in velocity
Put in numbers and find change in velocity
2007-02-20 22:52:11 · answer #2 · answered by hello 6 · 0⤊ 0⤋