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2 years ago

How efficient is a pulley system if an operator has to pull 2.5m of rope to lift a 250 N pail of water a distance of 4.5 m?

Physics

1 answer:

Alik [6]2 years ago

4 0

Answer:

the pulley system is very efficient

Explanation:

Given;

load, L = 250 N

distance moved by the load, a = 4.5 m

distance moved by the effort, e = 2.5 m

let the force or effort applied = F

Apply the principle of conservation of energy;

F x e = L x a

F = (L x a) / e

F = (250 x 4.5) / 2.5

F = 450 N

The efficient of the pulley system;

E = Output force / Input force

E = (450 x 2.5) / (250 x 4.5)

E = 1 = 100%

Therefore, the pulley system is very efficient

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liberstina [14]

(a) $Q = 1.70\times 10^{-2}\;\text{C}$ ;
(b) $V_\text{final} = 5.31\times 10^{2}\;\text{V}$ ;
(c) $E_\text{final} = 4.52\;\text{J}$ ;
(d) $\Delta E = 2.82\;\text{J}$ .

All four values are in 3 sig. fig.

<h3>Explanation</h3>

(a)

$Q = C\cdot V = 20.0\times 10^{-6} \times 850\;\text{V} = 1.70\times 10^{-2}\;\text{J}$ .

(b)

Sum of the final charge on the two capacitors should be the same as the sum of the initial charge. Voltage of the two capacitors should be the same. That is:

$C_1\cdot V_\text{final} +C_2 \cdot V_\text{final} = C_1\cdot V_\text{initial}$ ;

$(C_1+C_2)\cdot V_\text{final} = C_1\cdot V_\text{initial}$ ;

$\displaystyle V_\text{final} = \frac{C_1}{C_1+C_2}\cdot V_\text{initial}\\\phantom{V_\text{final}} = \frac{20.0\;\mu\text{F}}{20.0\;\mu\text{F} + 12.0\;\mu\text{F}} \times 850\;\text{V}\\\phantom{V_\text{final}} =531\;\text{V}$ .

(c)

$\displaystyle E = \frac{1}{2}\cdot C\cdot V^{2}$ .

$\displaystyle E_\text{final} = \frac{1}{2} (C_1 + C_2) \cdot {V_\text{final}}^{2} \\\phantom{E_\text{final}} = \frac{1}{2} \times (20.0\times 10^{-6} + 12.0\times 10^{-6}) \times 531.25\\\phantom{E_\text{final}} = 4.52\;\text{J}$ .

(d)

Initial energy of the system, which is the same as the initial energy in the $20.0\;\mu\text{F}$ capacitor:

$\displaystyle E_\text{initial} = \frac{1}{2} \times 20.0\times 10^{-6} \times 850^{2} = 7.225\;\text{J}$ .

Change in energy:

$\Delta E = 7.225\;\text{J} - 4.516\;\text{J} = 2.70\;\text{J}$ .

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