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. A 5.00-kg box slides across a rough horizontal floor, initially at 2.50 m/s. If friction brings

the box to rest after 1.50 s, determine the magnitude of the average rate in watts at which
friction dissipates the block’s mechanical energy.

User Isidor
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Answear: The magnitude of this average rate is 10.4 W.

Explanation:

To determine the magnitude of the average rate in watts at which friction dissipates the box's mechanical energy, we can use the work-energy principle, which states that the work done by all forces acting on an object is equal to the change in its mechanical energy.

The mechanical energy of the box initially is:

Ei = (1/2)mv^2 = (1/2)(5.00 kg)(2.50 m/s)^2 = 15.6 J

When the box comes to rest, its final mechanical energy is zero. Therefore, the work done by friction is:

W = Ef - Ei = -15.6 J

The negative sign indicates that the work done by friction is dissipative (i.e., it decreases the mechanical energy of the box).

The time taken for the box to come to rest is t = 1.50 s. Therefore, the average rate at which friction dissipates mechanical energy is:

P = W/t = -15.6 J / 1.50 s = -10.4 W
User Boedy
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