a. 1337.3 J work, in joules, is done by friction as the sled moved 28 m along the hill.
b.21,835 J work, in joules, is done by the rope on the sled this distance.
c. 23,170 J the work, in joules done by the gravitational force on the sled d. The net work done on the sled, in joules is 43,670 J.
<h3>What is friction work?</h3>
The work done by friction is the force of friction times the distance traveled times the cosine of the angle between the friction force and displacement
a. How much work is done by friction as the sled moves 28m along the hill?
ans. We use the formula:
friction work = -µ.mg.dcosθ
= -0.100 * 97.5 kg * 9.8 m/s² * 28 m * cos 60
= -1337.3 J
-1337.3 J work, in joules, is done by friction as the sled moved 28 m along the hill.
b. How much work is done by the rope on the sled in this distance?
We use the formula:
Rope work = -m.g.d(sinθ - µcosθ)
rope work = - 97.5 kg * 9.8 m/s² * 28 m (sin 60 – 0.100 * cos 60)
= 26,754 (0.816)
= 21,835 J
21,835 J work, in joules, is done by the rope on the sled this distance.
c. What is the work done by the gravitational force on the sled?
By using the formula:
Gravity work = mgdsinθ
= 97.5 kg * 9.8 m/s² * 28 m * sin 60
= 23,170 J
23,170 J the work, in joules done by the gravitational force on the sled .
D. What is the total work done?
By adding all the values
work done = -1337.3 + 21,835 + 23,170
= 43,670 J
The net work done on the sled, in joules is 43,670 J.
Learn more about friction work here:
brainly.com/question/14619763
#SPJ1
. Thus, the equation for velocity is 
.