# A 3.00-kg block starts from rest at the top of a 24.5° incline and slides 2.00 m down the incline in 1.35 s. Find the coefficient of kinetic friction

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### 2 Answers

A block with a mass of 3 kg starts from rest at the top of a 24.5 degree incline and slides 2 m down the incline in 1.35 s.

The distance moved by a body in t seconds that starts with a velocity u an travels under an acceleration a is given by D = u*t + (1/2)a*t^2

In the problem the slope of the incline is 24.5 degrees. This gives the component of acceleration due to gravity along the incline as g*sin 24.5. The deceleration due to friction is equal to 3*9.8*cos 24.5*C/3 = cos 24.5*9.8*C where C is the coefficient of kinetic friction. This gives:

2 = 0*1.35 + (1/2)*(9.8*sin 24.5 - cos 24.5*9.8*C)*1.35^2

=> 4/(1.35^2) = (9.8*sin 24.5 - cos 24.5*9.8*C)

=> cos 24.5*9.8*C = (9.8*sin 24.5 - 4/1.35^2)

=> (4/1.35^2 - 9.8*sin 24.5)/9.8*cos 24.5

=> C = 0.2096

The coefficient of kinetic friction is 0.2096

Mass of block = m = 3.00 kg

Slope of incline with horizontal = i = 24.5 degrees

Initial speed = u = 0 m/s

Distance moved down in direction of incline = s = 2.00 m

Time to move this distance = t = 1.35 s

Let "a" be the acceleration of the block in the direction of incline

s = u*t + 1/2 * a * t^2

=> 2 = 0*1.35 + 1/2*a*1.35^2

=> 2 = 0.91*a

=> a = 2.20 m/s^2

Force with which the block is sliding down the incline = m*a = 3*2.20 = 6.60 NFroce of gavity in the direction of incline due to blocks mass = m*g*sin(i) = 3*9.1*sin(24.5) = 11.32 N

Drag due to friction = 11.32 - 6.60 = 4.72 N

Force on the Block in verticle direction to the incline due to gravity = m*g*cos(i) = 3*91*cos(24.5) = 24.84 N

**Coefficient of kinetic friction** = 4.72/24.84 = **0.19**