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?
1 answer:
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
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