A 63 kg boy and sled are pulled along level ground with a force of 577 N. The boy and sled
are moving at a constant velocity of 2.4 m/s. What is the size of the frictional force?

Answers

Answer 1

Answer:

The size of the frictional force is 577 N.

Explanation:

What is frictional force?

Frictional forces are exerted every time you have two surfaces sliding across one another. Friction depends on:

how rougher are the surfaces on contact;how harder you push one surface to the other.

There are two types of friction:

static friction [tex]f_s[/tex];dynamic (sliding) friction [tex]f_d[/tex].

Now let's attack the problem:

First, you have to write the data down:

[tex]\bullet \quad m= 63\,kg\\\\\bullet \quad F= 577\,N\\\\\bullet \quad v= 2.4\,m/s[/tex]

If the boy and the sled moves on constant velocity, Newton's First Law Of Motion states that the net force acting on them must be zero:

[tex]F_{net}=0[/tex]

Applying Newton's Second Law Of Motion, we get:

[tex]F-f_d=0[/tex]

which means that:

[tex]F=f_d[/tex]

In other words, the dynamic frictional force equals the force that moves the boy and sled. So we may write:

[tex]\boxed{f_d=577\,N}[/tex]

Conclusion: the size of the (dynamic) frictional force is 577 N.

Happy studying!

Brainly Team

A 63 Kg Boy And Sled Are Pulled Along Level Ground With A Force Of 577 N. The Boy And Sledare Moving

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Answers

[tex]\\ \sf\longmapsto No\;of\:moles=\dfrac{Given\:mass}{Atomic\:mass}[/tex]

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Answer:

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Explanation:

I hope that helps you

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#Batangas

[tex]\\ \tt\longmapsto d=\dfrac{49.07}{8}\times 100[/tex]

[tex]\\ \tt\longmapsto d=6.13(100)[/tex]

[tex]\\ \tt\longmapsto d=613km[/tex]

#Puerto Princesa

[tex]\\ \tt\longmapsto d=\dfrac{58.10}{8}\times 100[/tex]

[tex]\\ \tt\longmapsto d=7.26(100)[/tex]

[tex]\\ \tt\longmapsto d=726km[/tex]

#Davao

[tex]\\ \tt\longmapsto d=\dfrac{39.47}{8}\times 100[/tex]

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Answer:

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Answer:

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Explanation:

Habiba and selehadin have measured the voltage across a resistor to be 5.26V and the current flowing through in to be 0.41A they work out the resistance .

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Considering the definitipn of current, voltage, power, and Ohm's law,

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In this case, you know:

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Replacing:

[tex]R=\frac{5.26 V}{0.41 A}[/tex]

Solving:

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Answers

Explanation:

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Answers

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Calculating Height
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Answers

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[tex]\\ \sf\longmapsto h=\dfrac{0^2-15.5^2}{2(10)}[/tex]

[tex]\\ \sf\longmapsto h=\dfrac{240.25}{20}[/tex]

[tex]\\ \sf\longmapsto h=12.012m[/tex]

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Answers

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Explanation:

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