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A 5.0-kg block and a 4.0-kg block are connected by a 0.6 kg rod, as shown in the figure. The links between the blocks and the rod are denoted by A and B. A vertical upward force of magnitude F of magnitude 150 N is applied to the upper block. What magnitude force does each of the links A and B exert?

A 5.0-kg block and a 4.0-kg block are connected by a 0.6 kg rod, as shown in the figure-example-1
User Votive
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To determine the magnitude of the force exerted by each of the links A and B, we can use the principle of conservation of energy.

The force applied to the upper block creates a gravitational potential energy (mgh) which will be converted into kinetic energy (1/2mv^2) of the blocks when they start moving.

The initial potential energy of the system is:

Ui = m1gh1 + m2gh2 + m3gh3

The final kinetic energy of the system is:

Uf = (1/2)m1v1^2 + (1/2)m2v2^2

Since the net force on the system is zero and there is no external force acting on it,

Ui = Uf

Where:

m1= 5.0 kg (mass of upper block)

m2= 4.0 kg (mass of lower block)

m3= 0.6 kg (mass of rod)

g = 9.8 m/s^2 (acceleration due to gravity)

v1 = velocity of upper block

v2 = velocity of lower block

The force exerted by link A is the force exerted on the upper block by link A and the force exerted by link B is the force exerted on the lower block by link B.

F(A) = m1a = m1g = 5.0 kg * 9.8 m/s^2 = 49 N

F(B) = m2a = m2g = 4.0 kg * 9.8 m/s^2 = 39.2 N

So the force exerted by link A is 49 N, and the force exerted by link B is 39.2 N.

User Bret Copeland
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