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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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:

∑F_{x}=mg sin(25) - F_{f}

∑F_{y}=F_{n}-mg cos(25)

Use Newton's 2nd law for each: in the x-direction, ∑F_{x}=ma, while in the y-direction ∑F_{y}=0.

mg sin(25) - F_{f} = ma

F_{n}- mg cos(25) =0

Get the normal force: F_{n} = mg cos(25)

Then the force of kinetic friction is: F_{f} = µ F_{n} = µ mg cos(25)

Therefore,

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

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