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2 years ago

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A 3.0-kg mass and a 5.0-kg mass hang vertically at the opposite ends of a very light rope that goes over an ideal pulley. If the

masses are gently released, what is the resulting acceleration of the masses

1 answer:

AleksAgata [21]2 years ago

5 0

Answer:

acceleration = 2.4525‬ m/s²

Explanation:

Data: Let m1 = 3.0 Kg, m2 = 5.0 Kg, g = 9.81 m/s²

Tension in the rope = T

Sol: m2 > m1

i) for downward motion of m2:

m2 a = m2 g - T

5 a = 5 × 9.81 m/s² - T

⇒ T = 49.05‬ m/s² - 5 a Eqn (a)‬

ii) for upward motion of m1

m a = T - m1 g

3 a = T - 3 × 9.8 m/s²

⇒ T = 3 a + 29.43‬ m/s² Eqn (b)

Equating Eqn (a) and(b)

49.05‬ m/s² - 5 a = T = 3 a + 29.43‬ m/s²

49.05‬ m/s² - 29.43‬ m/s² = 3 a + 5 a

19.62 m/s² = 8 a

⇒ a = 2.4525‬ m/s²

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Answer:

$5.32\cdot 10^4 g$

Explanation:

First of all, we need to find the volume of the room, which is given by

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So here it is given as

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$r = \frac{a}{\sqrt2}$

$r = \frac{10nm}{\sqrt2} = 7.07 nm$

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$U_i = qV = 2*1.6 * 10^{-19} * (-0.815) = -2.6 * 10^{-19} J$

Now at the mid point of one of the side

Electrostatic potential is given as

$V = 2\frac{kQ}{r_1} + 2\frac{kQ}{r_2}$

here we know that

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Now work done in this process is given as

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$W = (-0.267 * 10^{-19}) - (-0.26 * 10^{-19}}$

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