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A 5.0-kg object is pulled along a horizontal surface at a constant speed by a 15-n force acting 20° above the horizontal. How much work is done by this force as the object moves 6.0 m?

Respuesta :

skyluke89

Answer:

84.6 J

Explanation:

The work done by the force is given by

[tex]W=Fd cos \theta[/tex]

where

W is the work done

F = 15 N is the force applied

d = 6.0 m is the displacement

[tex]\theta=20^{\circ}[/tex] is the angle between the force's direction and the displacement

Substituting the numbers into the equation, we find

[tex]W=(15 N)(6.0 m)cos 20^{\circ} =84.6 J[/tex]

dsdrajlin

A 5.0 kg object is pulled by a 15-N force acting 20 °C above the horizontal. After moving for 6.0 m, the work done is 85 J.

What is work?

In physics, work is the energy transferred to or from an object via the application of force along a displacement.

A 5.0 kg object is pulled by a 15-N force acting 20 ° above the horizontal. We can calculate the work done after the object moved 6.0 m using the following expression.

W = F × s × cosθ

W = 15 N × 6.0 m × cos 20° = 85 J

where,

  • W is the work.
  • F is the force.
  • s is the displacement.
  • θ is the angle between F and s.

A 5.0 kg object is pulled by a 15-N force acting 20 °C above the horizontal. After moving for 6.0 m, the work done is 85 J.

Learn more about work here: https://brainly.com/question/62183

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