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How fast would a 15 kg box accelerate?

How fast would a 15 kg box accelerate down a 25º snowy slope if the coefficient of kinetic friction were 0.15?

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Robert J. | Certified High School AP Calculus and Physics TeacherCertified High School AP Calculus and Ph...
4.6 4.6 (13 lesson ratings) (13)
Check Marked as Best Answer
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
Brad M. | STEM Specialist plus Business, Accounting, Investment & EditingSTEM Specialist plus Business, Accountin...
4.9 4.9 (233 lesson ratings) (233)
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 :)
Andre W. | Friendly tutor for ALL math and physics coursesFriendly tutor for ALL math and physics ...
5.0 5.0 (3 lesson ratings) (3)
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
∑Fy=Fn-mg cos(25)
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