A150 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?

The

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

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

2.67 m/s²  (nearest hundredth)

Ithink your answer is d