Answer:
15.C Mucus reduces friction
Let
M = the mass of the planet
n = the mass of the satellite.
r = the radius of the planet
When the satellite is at a distance r from the surface of the planet, the distance between the centers of the two masses is 2r.
The gravitational force between them is

where
G = the gravitational constant.
When the satellite is on the surface of the planet, the distance between the two masses is r.
The gravitational force between them is

Answer:
The work done is given by 742.5 J while the coefficient of kinetic friction between the block and the surface is 0.46.
<h3>What is the work done?</h3>
The work done is given by the use of the formula;
W = F * x
Where;
F = force applied
x = distance covered
W = 150 N * 4.95 m = 742.5 J
Now;
The coefficient of kinetic friction is given by;
μ = F/mg
μ = 150/ 33 * 9.8
μ = 0.46
Learn more about work done:brainly.com/question/13662169
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15,000J/.15% = 100,000 J
If you want to check,
Multiply 100,000 by .15 and you get 15,000 J