<span>Reducing the distance between them. In theory, also increasing the mass; but you can't really change the mass of an object. However, you can compare the forces if you replace an object by a different object, which has a different mass.
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i hope this will work..
Answer:
a) variation of the energy is equal to the work of the friction force
b) W = Em_{f} -Em₀
, c) he conservation of mechanical energy
Explanation:
a) In an analysis of this problem we can use the energy law, where at the moment the mechanical energy is started it is totally potential, and at the lowest point it is totally kinetic, we can suppose two possibilities, that the friction is zero and therefore by equalizing the energy we set the velocity at the lowest point.
Another case is if the friction is different from zero and in this case the variation of the energy is equal to the work of the friction force, in value it will be lower than in the calculations.
b) the calluses that he would use are to hinder the worker's friction force and energy
W = Em_{f} -Em₀
N d = ½ m v² - m g (y₂-y₁)
y₂-y₁ = 35 -10 = 25m
c) if there is no friction, the physical principle is the conservation of mechanical energy
If there is friction, the principle is that the non-conservative work is equal to the variation of the energy
Answer:
a. 16 s b. -1.866 kJ
Explanation:
a. Since the initial rotational speed ω₀= 3313 rev/min = 3313/60 × 2π rad/s = 346.94 rad/s. Its rotational speed becomes ω₁ = 0.75ω₀ in time t = 4 s.
We find it rotational acceleration using α = (ω₁ - ω₀)/t = (0.75ω₀ - ω₀)/t = ω₀(0.75 - 1)/t = -0.25ω₀/t = (-0.25 × 346.94 rad/s)/4 s = -21.68 rad/s².
Since the turntable stops at ω = 0, the time it takes to stop is gotten from
ω = ω₀ + αt and t = (ω - ω₀)/α = (0 - 346.94 rad/s)/-21.68 rad/s² = (-346.94/-21.68) s = 16 s.
So it takes the turntable 16 s to stop.
b. The workdone by the turntable to stop W equals its rotational kinetic energy change.
So, W = 1/2Iω² - 1/2Iω₀² = 1/2 × 0.031 kgm² × 0² - 1/2 × 0.031 kgm² × (346.94 rad/s)² = 0 - 1865.7 J = -1865.7 J = -1.8657 kJ ≅ -1.866 kJ
