How fast would a 15 kg box accelerate down a 25º snowy slope if the coefficient of kinetic friction were 0.15?
How fast would a 15 kg box accelerate?
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Apply Newton's second law in the direction parallel to the incline,
mg sin25 - μmg cos25 = ma
Solve for a,
a = g(sin25 -0.15cos25) = 2.8 m/s^2 <==Answer
Hey Sun -- the 150N at 25 deg distributes 40% downhill and about 90% normal ...
the downhill 60N is opposed by 15% of the normal 135N or roughly 20N friction ...
the 40N downhill result moves the 15kg at nearly 2.7 m/s/s ... Best wishes :)
Draw a free-body diagram showing the three forces acting on the box: weight straight down, normal force perpendicular to the incline up, and friction along the incline opposite to the direction of motion.
Call the direction parallel to the incline the x-direction and the direction perpendicular to the incline the y-direction. Find the sum of all forces in the x- and y-direction:
∑Fx=mg sin(25) - Ff
Use Newton's 2nd law for each: in the x-direction, ∑Fx=ma, while in the y-direction ∑Fy=0.
mg sin(25) - Ff = ma
Fn- mg cos(25) =0
Get the normal force: Fn = mg cos(25)
Then the force of kinetic friction is: Ff = µ Fn = µ mg cos(25)
mg sin(25) - µ mg cos(25) = m a
Divide through by m, the answer is independent of the mass:
a = g sin(25) - µg cos(25)
a = 9.8 m/s² ( sin(25) - 0.15 cos(25) )
a = 2.8 m/s²