Question

A 15.0 m long plane is inclined at 30.0 degrees. If the coefficient of friction is 0.426, what force is required to move a 40.0 kg mass from rest at the bottom on the plane to the top of the plane with a final velocity of 8.00 m/s?

Answer 1

consider the motion of the mass parallel to the incline

v₀ = initial velocity at the bottom of incline = 0 m/s

v = final velocity at the top of incline = 8.00 m/s

a = acceleration

d = displacement = L = length of incline = 15 m

using the equation

v² = v²₀ + 2 a d

8² = 0² + 2 a (15)

64 = 30 a

a = 64/30

a = 2.13 m/s²

F = applied force

from the force diagram, perpendicular to incline , force equation is given as

N = mg Cos30

μ = Coefficient of friction = 0.426

frictional force acting on the mass is given as

f = μ N

f = μ mg Cos30

parallel to incline , force equation is given as

F - f - mg Sin30 = ma

F - μ mg Cos30 - mg Sin30 = ma

inserting the values

F - (0.426 x 40 x 9.8) Cos30 - (40 x 9.8) Sin30 = 40 (2.13)

F = 425.82 N