A 4.9 kg block slides down an inclined plane that makes an angle of 27◦ with the horizontal. Starting from rest, the block slides a distance of 2.7 m in 5.4 s. The acceleration of gravity is 9.81 m/s 2 . Find the coefficient of kinetic friction between the block and plane.

We define the x-axis in the direction parallel to the movement of the block on the inclined plane and the y-axis in the direction perpendicular to it.

Forces acting on the block

W: Weight of the block : In vertical direction

FN : Normal force : perpendicular to the inclined plane

fk : kinetic Friction force: parallel to the inclined plane

Calculated of the W

W= m*g

W= 4.9 kg* 9.8 m/s² = 48.02 N

x-y weight components

Wx = Wsin θ = 48.02*sin27° = 21.8 N

Wy = Wcos θ = 48.02*cos27° = 42.786 N

Calculated of the FN

We apply the formula (1)

∑Fy = m*ay ay = 0

FN - Wy = 0

FN = Wy

FN = 42.786 N

Calculated of the fk

fk = μk* FN= μk*42.786 Equation (1)

Kinematics of the block

Because the block moves with uniformly accelerated movement we apply the following formula to calculate the acceleration of the block :

d = v₀*t+(1/2)*a*t² Formula (2)

Where:

d:displacement (m)

v₀: initial speed (m/s)

t: time interval (m/s)

a: acceleration ( m/s²)

Data:

d= 2.7 m

v₀ = 0

t= 5.4 s

We replace data in the formula (2)

d = v₀*t+(1/2)*a*t²

2.7 = 0+(1/2)*a*( 5.4)²

2.7 = (14.58)*a

a = 2.7 / (14.58)

a= 0.185 m/s²

We apply the formula (1) to calculated μk:

∑Fx = m*ax , ax= a : acceleration of the block

Wx-fk= m*a , fk=μk*42.786 of the Equation (1)

21.8 - (42.786)*μk = (4.9)*(0.185)

21.8 -0.907= (42.786)*μk

20.89 = (42.786)*μk

μk = (20.89) / (42.786)

μk = 0.488