Omg. Ummm. I am pretty sure it is Shift toward the red end of the spectrum
Hi there!
To find the appropriate force needed to keep the block moving at a constant speed, we must use the dynamic friction force since the block would be in motion.
Recall:

The normal force of an object on an inclined plane is equivalent to the vertical component of its weight vector. However, the horizontal force applied contains a vertical component that contributes to this normal force.

We can plug in the known values to solve for one part of the normal force:
N = (1)(9.8)(cos30) + F(.5) = 8.49 + .5F
Now, we can plug this into the equation for the dynamic friction force:
Fd= (0.2)(8.49 + .5F) = 1.697 N + .1F
For a block to move with constant speed, the summation of forces must be equivalent to 0 N.
If a HORIZONTAL force is applied to the block, its horizontal component must be EQUIVALENT to the friction force. (∑F = 0 N). Thus:
Fcosθ = 1.697 + .1F
Solve for F:
Fcos(30) - .1F = 1.697
F(cos(30) - .1) = 1.697
F = 2.216 N
500J
Explanation:
Given parameters:
Weight of the body = 50N
Height = 10m
Unknown:
Work done = ?
Solution;
Work done is the force that moves a body through a particular distance in the direction of the force.
In this problem, we can solve the problem by relating work done to the potential energy used in lifting the mass.
The weight on a body, a force in the presence of gravity
weight(weight) = mg
where m = mass of body
g = acceleration due to gravity
Work done = Fxd
Where F = force on the body
d = distance moved
Potential energy = mgh
where h is the height
P.E = work done = Weight x height = 50 x 10 = 500J
Learn more:
Work done brainly.com/question/9100769
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