A box slides down a 30.0° ramp with an acceleration of 1.20 m/s^2. Determine the coefficient of kinetic friction between the box
and the ramp.
1 answer:
m = mass of the box
N = normal force on the box
f = kinetic frictional force on the box
a = acceleration of the box
μ = coefficient of kinetic friction
perpendicular to incline , force equation is given as
N = mg Cos30 eq-1
kinetic frictional force is given as
f = μ N
using eq-1
f = μ mg Cos30
parallel to incline , force equation is given as
mg Sin30 - f = ma
mg Sin30 - μ mg Cos30 = ma
"m" cancel out
a = g Sin30 - μ g Cos30
inserting the values
1.20 = (9.8) Sin30 - (9.8) Cos30 μ
μ = 0.44
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Answer:
0.56
Explanation:
Let the coefficient of friction is μ.
m = 4.3 kg, θ = 30 degree, initial velocity, u = 0, s = 2.7 m, t = 5.8 s
By the free body diagram,
Normal reaction, N = mg Cosθ = 4.3 x 9.8 x Cos 30 = 36.49 N
Friction force, f = μ N = 36.49 μ
Net force acting on the block,
Fnet = mg Sinθ - f = 4.3 x 9.8 x Sin 30 - 36.49 μ
Fnet = 21.07 - 36.49μ
Net acceleartion, a = Fnet / m
a = (21.07 - 36.49μ) / 4.3
use second equation of motion
s = ut + 1/2 a t^2
2.7 = 0 + 1/2 x (21.07 - 36.49μ) x 5.8 x 5.8 / 4.3
By solving we get
μ = 0.56
