A 50.0 kg object is in a 30 degree frictionless inclined plane that is exactly 1
ID: 1658578 • Letter: A
Question
A 50.0 kg object is in a 30 degree frictionless inclined plane that is exactly 100.0 meters long (as measured along the surface of the plane, not along the ground.) 1) How long will it take the object to slide to the ground from the top of the plane? 2) What is the normal force? 3) If the plane has a frictional coefficient of .200, how long will it take for the object to slide to the ground? 4) what is the minimal frictional coefficient necessary on the plane for the object not to slide? A 50.0 kg object is in a 30 degree frictionless inclined plane that is exactly 100.0 meters long (as measured along the surface of the plane, not along the ground.) 1) How long will it take the object to slide to the ground from the top of the plane? 2) What is the normal force? 3) If the plane has a frictional coefficient of .200, how long will it take for the object to slide to the ground? 4) what is the minimal frictional coefficient necessary on the plane for the object not to slide?Explanation / Answer
here,
mass , m = 50 kg
theta = 30 degree
l = 100 m
a)
accelration , a = g * sin(theta) = 4.9 m/^2
the time taken be t
l = 0 + 0.5 * a * t^2
100 = 0 + 0.5 * 4.9 * t^2
t = 6.39 s
b)
the normal force , N = m * g * cos(theta)
N = 50 * 9.81 * cos(30) N
N = 424.8 N
c)
uk = 0.2
accelration , a = ( g * sin(theta) - uk * g * cos(theta))
a = ( 9.81 * sin(30) - 0.2 * 9.81 * cos(0.866)) m/s^2
a = 3.2 m/s^2
let the time taken be t
l = 0 + 0.5 * a * t^2
100 = 0 + 0.5 * 3.2 * t^2
t = 7.9 s
d)
let the coefficient of friction be us
equating the forces
m * g * sin(theta) - us * m * g * cos(theta) = 0
us = tan(theta ) = 0.58
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