Question

A 44 kg block is moving at 6.5 m/s at the bottom of a rough 22 degree incline. if the coefficient of friction on the incline is 0.16, what distance along the incline does the block travel?

Answer 1

we will divide the forces into two components first Normal to the inclined and second parallel to the inclined.

Summing the force Normal to inclined plane,

- F_normal + m*g* cos ß = 0

F_normal = m*g* cos ß -----> (1)

Now, forces parallel to the inclined,

- F_friction + m*g* sin ß = m*a ---->(2)

we know that,

F_friction = μ* F_normal ------>(3)

using (1) in (3),

- F_friction = μ*m*g* cos ß ----->(4)

using (4) in (2) and using known values,

0.16*m*g* cos (22) + m*g* sin(22) = m*a

a = [sin (22) + 0.16* cos (22)]*g

solving for a,

a = - 5.124 m/s²

using third equation of motion,

V² - U² = 2*a*s

0² - (6.5)² = 2*(-5.124)*s

solving for s,

s = 4.12 m

Thus the block will travel to 4.12 m on the inclined plane.