How fast would a 15 kg box accelerate?

Question

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

Answer 1

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

Answer 2

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 :)

Answer 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:

∑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