Answer:
part (a). 176580 J
part (b). 197381 J
Explanation:
Given,
- Density of the chain =

- Length of the chain = L = 60 m
- Acceleration due to gravity = g = 9.81

part (a)
Let dy be the small element of the chain at a distance of 'y' from the ground.
mass of the small element of the chain = 
Work done due to the small element,

Total work done to wind the entire chain = w

part (b)
- mass of the block connected to the chain = m = 35 kg
Total work done to wind the chain = work done due to the chain + work done due to the mass

According to the Jefferson lab, "The scientific definition of work is: using a force to move an object a distance (when both the force and the motion of the object are in the same direction.)"
Answer:
First of all the formula is F= uR,( force= static friction× reaction)
mass= 5+25=30
F= 50
R= mg(30×10)=300
u= ?
F=UR
u= F/R
u= 50/300=0.17N
Answer:
The function has a maximum in 
The maximum is:

Explanation:
Find the first derivative of the function for the inflection point, then equal to zero and solve for x




Now find the second derivative of the function and evaluate at x = 3.
If
the function has a maximum
If
the function has a minimum

Note that:

the function has a maximum in 
The maximum is:
