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77julia77 [94]
3 years ago
5

So far, you’ve been working with an "ideal" pulley system. How do you think real pulley systems are different, and how would tha

t affect the mechanical advantage of real pulley systems?
Plz really need help with this. And thank you so much
Physics
1 answer:
almond37 [142]3 years ago
8 0

Answer:

In an ideal pulley system is assumed as a perfect system, and the efficiency of the pulley system is taken as 100% such that there are no losses of the energy input to the system through the system's component

However, in a real pulley system, there are several means through which energy is lost from the system through friction, which is converted into heat, sound, as well as other forms of energy

Given that the mechanical advantage = Force output/(Force input), and that the input force is known, the energy loss comes from the output force which is then reduced, and therefore, the Actual Mechanical Advantage (AMA) is less than the Ideal Mechanical Advantage of an "ideal" pulley system

The relationship between the actual and ideal mechanical advantage is given by the efficiency of the pulley system as follows;

Efficiency \, \% = \dfrac{AMA}{IMA}  \times 100

Explanation:

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assuming ideal gas behaviour of the gas , the equation for ideal gas is

P*V=n*R*T

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P = absolute pressure

V= volume

T= absolute temperature

n= number of moles of gas

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P=n*R*T/V

the work that is done by the gas is calculated through

W=∫pdV=  ∫ (n*R*T/V) dV

for an isothermal process T=constant and since the piston is closed vessel also n=constant during the process then denoting 1 and 2 for initial and final state respectively:

W=∫pdV=  ∫ (n*R*T/V) dV =  n*R*T  ∫(1/V) dV = n*R*T * ln (V₂/V₁)

since

P₁=n*R*T/V₁

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dividing both equations

V₂/V₁ = P₁/P₂

W= n*R*T * ln (V₂/V₁)  = n*R*T * ln (P₁/P₂ )

replacing values

P₁=n*R*T/V₁ = 2 moles* 8.314 J/mol K* 300K / 0.1 m3= 49884 Pa

since P₂ = 1 atm = 101325 Pa

W= n*R*T * ln (P₁/P₂ ) = 2 mol * 8.314 J/mol K * 300K * (49884 Pa/101325 Pa) = -3534.94 J

5 0
3 years ago
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