a 15kg box is released on a 32 degress incline and accerlerates at 0.30 ms2 what is the friction force and the coefficent of kinetic friction
2007-11-04 07:33:17 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Physics
i have no clue. look it up or ask your teacher for help
2007-11-04 07:35:48 · answer #1 · answered by Anonymous · 0⤊ 0⤋
The plane is inclined at 32 degrees (from the horizontal, I guess).
The force of gravity (15 kg * 9.8 m/s = 147 N) is directed towards the centre of the Earth (i.e., exactly vertical).
This force can be broken down into its two constituents in relation to the inclined plane.
The portion directly perpendicular to the plane is responsible for the friction. It is equal to 147 N * Cos(32) and the dynamic friction is 147 N * Cos(32) * k where k is the coefficient of kinetic friction. Here, we assume (to make calculations easier) that this coefficient does not change with speed (which is often a reasonable assumption over small ranges of speeds).
The force pulling the box forward along the plane is 147 N * Sin(32).
Some of this force accelerates the box and the rest goes to combat the friction.
The portion that accelerates the box can be found with
F = m a
m is the 15 kg of the box
a is the given 0.3 m/s^2
15 kg * 0.3 m/s^2 = 4.5 N
The difference
[ 147*Sin(32) ] - 4.5
is what was used up to combat the force of friction
(i.e., it must be equal to the friction force)
Now that you have the friction force, you can find the coefficient of kinetic friction k, using the equation
Friction Force = 147 N * Cos(32) * k
2007-11-04 15:45:45 · answer #2 · answered by Raymond 7 · 0⤊ 0⤋
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im studying on newton's laws and friction stuff..im not good at the math part..lol
2007-11-04 15:36:09 · answer #3 · answered by yesenia 2 · 0⤊ 0⤋