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A boxed 10 kg computer monitor is dragged by friction 5.5 m up along the moving surface of a conveyor belt inclined at an angle of 36.9° above the horizontal. If the monitor’s speed is a constant, how much work is done on the monitor by friction?

Respuesta :

Answer:

324 J

Explanation:

58.84 * 5.5 J = 324 J

I got 58.84 by 10 * 9.8 (gravity) * sin(36.9)

Lanuel

The amount of work done on the monitor by friction is equal to 431 Joules.

Given the following data:

  • Mass of computer monitor = 10 kg
  • Distance = 5.5 meters
  • Angle of inclination = 36.9°

To determine the amount of work done on the monitor by friction:

Mathematically, work done is given by the formula:

[tex]Work\;done = mgdcos\theta[/tex]

Where:

  • m is the mass of an object.
  • g is the acceleration due to gravity.
  • d is the distance.

Substituting the given parameters into the formula, we have;

[tex]Work\;done = 10 \times 9.8 \times 5.5 \times cos(36.9)\\\\Work\;done = 539 \times 0.7997[/tex]

Work energy = 431 Joules.

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