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A warehouse worker pushes a 19.1 kg box, increasing its speed from rest to 3.40 m/s over a period of 0.933 seconds. How much power did the worker produce?

Select one:
A. 118 W
B. 110 W
C. 126 W
D. 103 W

Answer :

Answer:

a. 118 W

Explanation:

Power is rate of doing work.

We must ASSUME that friction is not relevant as we are given no clue as to its values. We must also ASSUME the surface is horizontal.

The work done changes the kinetic energy

W = ½m(v₁² - v₀²) = ½(19.1)(3.40² - 0²) = 110 J

as this occurs in 0.933 s, power is

P = 110 / 0.933 = 118 W(atts)

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Rewritten by : Barada

Final answer:

The worker produced power approximately equal to 118 Watts. This was calculated using the formula for power in physics, taking into account the work done and the time taken for it.

Explanation:

The formula for power in physics is given as Power = Work / Time. Here we need to first calculate the Work done.

The Work done can be calculated using the formula: Work = 0.5 * mass * (final speed)^2 - 0.5 * mass * (initial speed)^2. In this case, as the initial speed was at rest, it is zero, so, the work done would be Work = 0.5 * 19.1 kg * (3.40 m/s)^2 = 109.594 Joules.

Next, using the work done and the time taken, we can calculate power.

Power = Work / Time = 109.594 Joules / 0.933 seconds = 117.46 Watts. Thus the closest answer would be 118 Watts.

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