Need help finding coefficient of kinetic friction.?

Question

Stuck on another physics question.

Ice blocks slide with an acceleration of 1.22 m/s^2 down a chute at an angle of 12 degrees below the horizontal. What is the coefficient of kinetic friction between the ice and chute?

I don't know how to solve this since I'm only given the acceleration and degree..

THANKS

My book says that the answer is 0.085 -- i have no idea how this was achieved..

Answer 1

In the absence of friction, the acceleration of an object down a 12-degree slide would be

9.81 * sin 12, or around 2.04

Instead, your obejct is accelerating at 1.22.

The loss of acceleration due to friction is

(9.81 * sin 12) - 1.22

and the coefficient of friction is thus

[(9.81 * sin 12) - 1.22] / (9.81 * sin 12)

Answer 2

Draw a free body diagram of the block with the axis tiltled 12 degrees from the horizontal..(i drew it such that the ice is moving to the right)
be careful of the Freebody diagram.. its a bit tricky knowing the angles

summing forces along y we get

N = mgcosθ

summing forces along x we get

mgsinθ - μN = 1.22m
*substitute N and cancel out all the masses
gsinθ - μgcosθ = 1.22
μ = (gsinθ - 1.22)/gcosθ
μ = (9.81sin12 - 1.22)/9.81cos12

μ = 0.085415325

if you have questions about the FBD, just YM or mail me (only once though) :)

Answer 3

Draw the free body diagram and equate the forces along the two perpendicular axis which u have chosen in your frame of reference. The equation comes out to be
mgsin(x)-(mu)N=ma, where N=mgcos(x)
finally, (mu)=1/cos(x)[sin(x)-a/g]
putting the respective values u get (mu) as 0.085.