Refer to the diagram shown below.
θ = 28°, the angle of the slope.
W = 150 N, the weight of the sled.
The normal reaction is
N = W cosθ = 150*cos(28) = 132.442 N
The component of the weight acting down the plane is
F = W sin θ = 150*sin(8) = 70.421 N
Case A; The sled is pulled up the slope.
Because the sled is in dynamic equilibrium, therefore
F + μN = 100
where μ = the kinetic coefficient of friction.
That is,
70.421 + 132.442μ = 100
μ = (100 - 70.421)/132.442 = 0.2233
Case B: The sled accelerates down the slope.
When the applied force of 100 N is removed, the sled accelerates down the slope due to its weight acting down the slope. The motion is opposed by kinetic friction.
That is,
Wsinθ - μWcosθ = ma
where
m = (150 N)/(9.8 m/s²) = 15.306 kg, the mass of the sled
a = the acceleration of the sled
Therefore
70.421 - 0.2233*132.442 = 15.306*a
a = 40.8467/15.306 = 2.669 m/s²
Answer:2.67 m/s² (nearest hundredth)
