To solve this problem we must resort to the Work Theorem, internal energy and Heat transfer. Summarized in the first law of thermodynamics.

Where,
Q = Heat
U = Internal Energy
By reference system and nomenclature we know that the work done ON the system is taken negative and the heat extracted is also considered negative, therefore
Work is done ON the system
Heat is extracted FROM the system
Therefore the value of the Work done on the system is -158.0J
Answer:
Part a)

Part b)

Explanation:
As we know that by parallel axis theorem we will have

Part a)
here we know that for a stick the moment of inertia for an axis passing through its COM is given as

now if we need to find the inertia from its end then we will have



Part b)
here we know that for a cube the moment of inertia for an axis passing through its COM is given as

now if we need to find the inertia about an axis passing through its edge



Answer:
a) F = 4.9 10⁴ N, b) F₁ = 122.5 N
Explanation:
To solve this problem we use that the pressure is transmitted throughout the entire fluid, being the same for the same height
1) pressure is defined by the relation
P = F / A
to lift the weight of the truck the force of the piston must be equal to the weight of the truck
∑F = 0
F-W = 0
F = W = mg
F = 5000 9.8
F = 4.9 10⁴ N
the area of the pisto is
A = pi r²
A = pi d² / 4
A = pi 1 ^ 2/4
A = 0.7854 m²
pressure is
P = 4.9 104 / 0.7854
P = 3.85 104 Pa
2) Let's find a point with the same height on the two pistons, the pressure is the same
where subscript 1 is for the small piston and subscript 2 is for the large piston
F₁ = 
the force applied must be equal to the weight of the truck
F₁ =
F₁ = (0.05 / 1) ² 5000 9.8
F₁ = 122.5 N
The solution for the problem is:
Constant speed means Fnet = 0.
Let m = mass of wood block and Θ = angle of ramp; then if µk = 0.35 …
The computation would be:
Fnet = 0 = mg (sin Θ) - (µk) (mg) (cos Θ)
mg (sin Θ) = µk (mg) (cos Θ)
µk = tan Θ
Θ = arctan(µk)
= arctan (0.35)
≈ 19.3°