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A force of constant magnitude pushes a box up a vertical surface, as shown in the figure.

The box moves at a constant speed. If the mass of the box is 3.1 kg, it is pushed 2.5 m vertically upward, the coefficient of friction is 0.35, and the angle is 30.0°, determine the following.

A Force Of Constant Magnitude Pushes A Box Up A Vertical Surface As Shown In The Figure The Box Moves At A Constant Speed If The Mass Of The Box Is 31 Kg It Is class=

Sagot :

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The work done on the box by the applied force is zero.

The work done by the force of gravity is 75.95 J

The work done on the box by the normal force is 75.95 J.

The given parameters:

  • Mass of the box, m = 3.1 kg
  • Distance moved by the box, d = 2.5 m
  • Coefficient of friction, = 0.35
  • Inclination of the force, θ = 30⁰

What is work - done?

  • Work is said to be done when the applied force moves an object to a certain distance

The work done on the box by the applied force is calculated as;

[tex]W = Fd cos(\theta)\\\\W = (ma)d \times cos(\theta)[/tex]

where;

a is the acceleration of the box

The acceleration of the box is zero since the box moved at a constant speed.

[tex]W = (0) d \times cos(30)\\\\W = 0 \ J[/tex]

The work done by the force of gravity is calculated as follows;

[tex]W = mg \times d\\\\W = 3.1 \times 9.8 \times 2.5 \\\\W = 75.95 \ J[/tex]

The work done on the box by the normal force is calculated as follows;

[tex]W = (F_n) \times d\\\\W = (mg + F sin\theta) \times d\\\\W = (mg + 0) \times d\\\\W = mgd\\\\W = 3.1 \times 9.8 \times 2.5\\\\W = 75.95 \ J[/tex]

Learn more about work done here: https://brainly.com/question/8119756

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