Question

A 150 n sled is being pulled up a 28 ° rough ramp at constant speed by a force of 100 n parallel to the ramp. with what acceleration will the crate slide down, if it is released at some point on the ramp?

Answer 1

The answer:

we can use newton's second law for finding the value of acceleration

that is    F=MxA
M the mass
A is the acceleration

F= P-Fk,  Fk= -Mgcosθμ and P=Mgsinθ
μ is the coefficient of friction

The value of μ can be found with the above condition
(pulled up a 28 ° rough ramp at constant speed)
this implies the acceleration is equal to zero
 100N - mgsin(28°) -mgcos(28°)μ = 0 , from where 
μ= [100N - mgsin(28°)] / mgcos(28°)
μ=0.2

and then, 
F=MxA= Mgsinθ -Mgcosθμ   finally  A=gsin28° -gcos28° x0.2 =2.6m/s²

Answer 2

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)