A 42 kg of ice slides down the plane incline 34 degrees in the figure. Assuming friction is negligible, what is the acceleration of the block down the incline? If the kinetic coefficient of friction is 0.060, what is the acceleration of the block down the incline?
Please show and explain all work. Please show the formula or formulas that were used.
2007-12-15 04:52:09 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Physics
Orient your coordinate system such that the y-axis is normal to the plane and the x-axis parallel to it. Then contruct a free-body diagram for the mass...
Normal force N, along the positive y-axis
Friction force μN, along the negative x-axis
Gravity mg, straight down, so it has components
mg sin 30 on the positive x-axis and mg cos 30 on the negative y axis.
Using Newton's Second Law on the y-axis we get N = mg cos 30. Then the net force on the x-axis is
F = ma = mg sin 30 - μ mg cos 30
The acceleration is thus g sin 30 - μg cos 30.
2007-12-15 04:57:19 · answer #1 · answered by jgoulden 7 · 1⤊ 0⤋
F=ma
Driving force - Fricton = ma
Friction = µ(mgcosx)
mgsinx - µ(mgcosx) = ma
42(9.8)sin34 - 0.060(42)(9.8)cos34 = 42a
a = 4.992616361 m/s²
2007-12-15 05:37:22 · answer #2 · answered by Murtaza 6 · 1⤊ 0⤋