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340 kg box is lifted to a height of 1.75 m, determine the amount of work done. if the time used to lift the box is 0.5 minute, determine the power.

Sagot :

[tex]\tt{\huge{\blue{Explanation:}}}[/tex]

The power of an object doing a work W in a given time t is given by

[tex]\boxed{P = \frac{W}{t}}[/tex]

where:

P = power

W = work

t = time

[tex]\tt{\huge{\red{Solution:}}}[/tex]

Solving for the force

F = weight = mg

F = (340 kg)(9.8 m/s²)

F = 3,332 N

Solving for the work

W = Fd

W = (3,332 N)(1.75 m)

W = 5,831 J

Solving for the power

Note: 0.5 min = 30 s

[tex]P = \dfrac{W}{t}[/tex]

[tex]P = \dfrac{\text{5,831 J}}{\text{30 s}}[/tex]

[tex]\boxed{P = \text{194.37 W}}[/tex]

Therefore, the power is 194.37 W.

[tex]\\[/tex]

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