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3 years ago

It is easier to lift the same load by using three pulley system than by using two-pulley system. why give reason.

Physics

1 answer:

tresset_1 [31]3 years ago

7 0

<h2>QUESTION:- It is easier to lift the same load by using three pulley system than by using two-pulley system.</h2>

<h2>ANSWER:- IN CASE OF IDEAL PULLEY SYSTEM</h2>

<h2>REASON:- </h2>

Logic behind is lies behind the mechanical advantage of the provided bt the Pulley system.

as if we calculate the mechanical advantage of the 2 Pulley system we will have the value 2

And if we will calculate the mechanical advantage of the 3 pulley system then we will get the value of 3

so due to extra mechanical advantage we feel it easy to move with 3 pulley system then 2 Pulley system

$\red \star{Thanks \: And \: Brainlist} \blue\star \\ \green\star If \: U \: Liked \: My \: Answer \purple \star$

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3 years ago

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Answer:

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(b) V (A) = 0.7 m/s,

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Explanation:

Given:

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Sol:

a) law of conservation of momentum

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so 3.50 Kg x 2.00 m/s + 6.50 Kg x 0 m/s = (3.50 Kg + 6.50 Kg ) V

after solving V = 0.7 m/s

After the collision the velocities of the both block will be as the the spring is compressed maximum.

V (A) = 0.7 m/s

b) V(A) = 0.7 m/s ( Part (a) and Part (a) are repeated )

c) as stated above the in the Part (a)

V(B) = 0.7 m/s

d) When the both blocks moved apart after the collision:

Let u=velocity of block A after the collision.

and v = velocity of block B after the collision.

then conservation of momentum

M(a) x V(A) + M(B) x V(B) = M(a) x v + M(B) x u

⇒ 3.50 Kg x 2.00 m/s + 6.50 Kg x 0 m/s = 3.50 Kg x u + 6.50 Kg x v

⇒ 2.00 m/s = u + 1.86 v -----eqn (1) ( dividing both side by 3.50 Kg)

For elastic collision

the velocity relative approach = velocity relative separation

so 2.00 m/s = v-u ----- eqn (2)

⇒v = u + 2.00 m/s

putting this value in eqn (1) we get

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e) putting v= 2.00 m/s in eqn (1)

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3 years ago

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In the given question:

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3 years ago

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Answer:

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Before (b) = After (a)

$m_{1}v_{1b} + m_{2}v_{2b} = m_{1}v_{1a} + m_{2}v_{2a}$

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<u>Energy:</u>

Before (b) = After (a)

$\frac{1}{2}m_{1}v_{1b}^{2} + \frac{1}{2}m_{2}v_{2b}^{2} = \frac{1}{2}m_{1}v_{1a}^{2} + \frac{1}{2}m_{2}v_{2a}^{2}$

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<u>Now, by entering equation (3) into (2) we have: </u>

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I hope it helps you!

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3 years ago

It is easier to lift the same load by using three pulley system than by using two-pulley system. why give reason.​

It is easier to lift the same load by using three pulley system than by using two-pulley system. why give reason.