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3 years ago

A box weighing 18 N requires a force of 6.0 N to drag it at a constant rate. What is the coefficient of sliding friction?

Physics

2 answers:

Rzqust [24]3 years ago

6 0

0.33 . Equation is Force of friction equals normal force times coefficient of friction, so 6=18u. Divide 6 by 18

kolezko [41]3 years ago

3 0

Answer:

Coefficient of sliding friction is 0.33.

Explanation:

The force of sliding friction is proportional to the normal force. It is given by:

$F_s=\mu _sF_N$

The normal force of the box is equal to its weight as the force in the vertical direction is balanced.

$F_N= 18N$

$F_s= 6.0N$ (given)

Thus, the coefficient of sliding friction is:

$\mu _s=\frac{F_s}{F_N}=\frac{6.0}{18} = 0.33$

Thus, the coefficient of sliding friction is 0.33.

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Explanation:

The computation of the velocity of the existing water is shown below:

Data provided in the question

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Based on the above information, first we have to determine the height which is

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Now applied the Bernoulli equation to section 1 and 2

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Now

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hope it hepls

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