Question

A 71.9 Kg mass is placed on an incline plane that is 38 degrees from the horizontal.

Assuming friction is present and the coefficient of friction is 0.61, what is the acceleration of
the mass? If the distance of the incline plane is 4.3 m, how long does it take the mass to
reach the bottom of the plane taking friction into account?

I need the steps on how to solve the problem please. I appreciate your help

Answer 1

acceleration, a = 11m/s² ; time, t = 0.84sec

Step-by-step explanation:

Fw = mass(m) × acceleration due to gravity(g)

Fw = 71.9kg × 10m/s²

Fw = 719N

F = Fw(sintheta+coefficient of friction(costheta))

F = 719(sin38° + 0.61(cos38°)

F = 719(0.615661475+0.480686559)

F = 719(1.096348034)

F = 788.27N

Force(F) = mass(m) × acceleration(a)

Make acceleration the subject

acceleration = Force÷ mass

a = 788.27÷719

a = 11m/s²

using this formula v² = u² +2gh

v² = 0² + 2×10m/s²×4.3m

√v² = √86

v = 9.27m/s

acceleration = velocity/time

make time the subject

time = velocity/acceleration

t = 9.27m/s/11m/s²

t = 0.84sec