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What power output is required of a motor that needs to lift a 450 kg elevator car 12 meters in 8.0 seconds?\

Sagot :

MrRoyal

Answer:

[tex]Power = 6.62kw[/tex]

Explanation:

Given

[tex]Mass = 450kg[/tex]

[tex]Height = 12m[/tex]

[tex]Time = 8s[/tex]

Required

Determine the power output required

First, we calculate the energy used to lift the object.

To do this, we make use of potential energy formula:

[tex]Energy = mgh[/tex]

Where

[tex]g = 9.81m/s^2[/tex]

So, we have:

[tex]Energy = 450 * 9.81 * 12[/tex]

[tex]Energy = 52974J[/tex]

The power output is then calculated as:

[tex]Power = \frac{Energy}{Time}[/tex]

[tex]Power = \frac{52974}{8}[/tex]

[tex]Power = 6621.75W[/tex]

Convert to kilowatt

[tex]Power = \frac{6621.75kW}{1000}[/tex]

[tex]Power = 6.62175kW[/tex]

[tex]Power = 6.62kw[/tex] --- approximated

Hence, the power output is 6.62kw

fichoh

Using the power formula, the required power output needed to lift the load is 6.615 Kw

Given the Parameters :

  • Mass of car = 450 kg
  • Distance = 12 meters
  • Time taken = 8 seconds

Recall :

  • Power = (Workdone ÷ Δtime)

  • Workdone = Force × Distance ;
  • Force = mg ;
  • g = Acceleration due the gravity, = 9.8 m/s²

Hence,

Power = [(mgh ÷ t)]

Power = [(450×9.8×12) ÷ 8]

Power = 52920 ÷ 8

Power = 6615 watt

Hence, the power output required is 6615 watts = 6.615 Kw

Learn more :https://brainly.com/question/17081810

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