Answered

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Block Y with mass m, falls onto and sticks to block X, which is attached to a vertical spring, as shown in Figure 1. A short time later, as shown in Figure 2. the blocks are momentarily at rest. At that moment, block Y exerts a force of magnitude F on block X, and block X exerts a force of magnitude Fon block Y. Which of the following correctly relates F F and myg at the instant shown in Figure 2?
(A) (Fup= Fdown)>mg
(B) (Fup=mg) > Fdown
(C) mg>Fup > Fdown
(D) Fup = Fdown = mg

Block Y With Mass M Falls Onto And Sticks To Block X Which Is Attached To A Vertical Spring As Shown In Figure 1 A Short Time Later As Shown In Figure 2 The Blo class=

Sagot :

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The formula that correctly relates the force and weight of the blocks is Fup = Fdown = mg.

Forces in equilibrium

The net force at a given equilibrium position is zero.

∑F = 0

According Newton's third law of motion, action and reaction are equal and opposite. That is, the force acting on an object is equal to the reaction exerted by the object.

Fa = -Fb

Thus, at the given equilibrium position, the force exerted by block Y on X is equal to the force exerted by block X on Y.

The formula that correctly relates the force and weight of the blocks is Fup = Fdown = mg.

Learn more about Newton's third law of motion here: https://brainly.com/question/25998091

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