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Vikentia [17]
3 years ago
11

A box is pulled 6 meters across the ground at a constant velocity by a horizontally applied force of 50 newtons. At the same tim

e, kinetic friction acts on the box as it slides so that it eventually stops when the pulling force stops. What is the magnitude of the frictional force? How much work does the frictional force do? What is the net work done on the box?
Physics
1 answer:
Vikentia [17]3 years ago
7 0

Answer:

(a) Friction force = 50 N

(b) Work done by friction = 300 j

(c) Net work done = 0 j

Explanation:

We have given that the box is pulled by 6 meter so d = 6 m

Force applied on the box F = 60 N

We have have given that velocity is constant so acceleration will be zero

So to applied force will be utilized in balancing the friction force

So friction force F_{friction}=50N

Work done by friction force W_{friction}=F_{friction}\times d=50\times 6=300j

Work done by applied force W=F\times d=50\times 6=300j

So net work done = 300-300 = 0 j

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The period of rotation of Mars is 1 day and 37 minutes. Determine its frequency of rotation in Hertz.
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The frequency of rotation of Mars is 0.0000113 Hertz.

<u>Given the following data:</u>

  • Period = 1 day and 37 minutes.

To find the frequency of rotation in Hertz:

First of all, we would convert the the value of period in days and minutes to seconds because the period of oscillation of a physical object is measured in seconds.

<u>Conversion:</u>

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Cross-multiplying, we have:

X = 60 × 1477

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Now, we can find the frequency of rotation of Mars by using the formula:

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Therefore, the frequency of rotation of Mars is 0.0000113 Hertz.

Read more: brainly.com/question/14708169

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