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Nana76 [90]
3 years ago
7

A force of 48 newtons is required to start a 5.0 kg box moving across a horizontal concrete floor. What is the coefficient of st

atic friction between the box and the floor? What is the coefficient of kinetic friction when the acceleration is 0.70 m/s/s?
Physics
1 answer:
Soloha48 [4]3 years ago
7 0

1) The coefficient of static friction is 0.980

2) The coefficient of kinetic friction is 0.908

Explanation:

1)

In the first situation, the box is still at rest. There are two forces acting on the box:

- The force of push, F, forward

- The force of static friction, F_f

Since the box is in equilibrium we have

F_f = F (1)

The value of frictional force changes from zero to a maximum value which is given by:

F_f = \mu_s mg (2)

where

\mu_s is the coefficient of static friction

m is the mass of the box

g is the acceleration of gravity

So, if the force F needed to put the box in motion is 48 N, it means that this is also the maximum value of the force of friction. So, we can combine eq.(1) and (2) to find the coefficient of static friction:

\mu_s = \frac{F}{mg}

where:

F = 48 N

m = 5.0 kg

g=9.8 m/s^2

Substituting,

\mu_s = \frac{48}{(5.0)(9.8)}=0.980

2)

In this second situation, the object is already in motion, so the equation of motion is:

F-F_f = ma

where

F = 48 N is the force applied forward

a = 0.70 m/s^2 is the acceleration of the box

F_f = \mu_k mg is the force of kinetic friction, where

\mu_k is the coefficient of kinetic friction

We can therefore rearrange the equation to find the coefficient:

F-\mu_k mg = ma\\\mu_k mg = F-ma\\\mu_k = \frac{F-ma}{mg}=\frac{48-(5.0)(0.70)}{(5.0)(9.8)}=0.908

Learn more about friction:

brainly.com/question/6217246

brainly.com/question/5884009

brainly.com/question/3017271

brainly.com/question/2235246

#LearnwithBrainly

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<h3 /><h3 /><h3> Kinetic enrgy.</h3>

Kinetic energy is the energy due to the movement of bodies, it is given by the relation

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In a compound motion it is common to separate energy into parts to simplify calculations.

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          K_{earth} = \frac{1}{2} \  m r^2_{tras} w^2_{tras} + \frac{1}{2} ( \frac{2}{5} m r^2_{earth}) w^2_{rot}  

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Let's substitute.

        K_{earth} = \frac{1}{2} m ( \frac{2\pi r^2_{tras}}{T_{tras}})^2  \ + \frac{1}{5} m (\frac{2\pi r^2_{earth} }{T^2_{rot}})^2  

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Let's calculate.

        K_{earth} = 4 \pi^2 5.98 \ 10^{24}  ( \frac{1}{2} ( \frac{1.496 \ 10^{11}}{3.15 \ 10^7 } )^2  \ +  \frac{1}{5}( \frac{6.37 \ 10^6 }{8.64 \ 10^4})^2 )

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Let's analyze the kinetic energy for the Sun, this is inside the solar system therefore it has no translation movement and is approximately a sphere with a rotation period of T_{Sum} = 27 days.

The kinetic energy of the sun is;

          K_{sum} = K_{rot} =  \frac{1}{2} I w^2  

          K_{sum} = \frac{1}{2} (\frac{2}{5} M R^2) (\frac{2\pi}{T_{sum}})^2  

          K_{sum} = \frac{4\pi^2 }{5} M (\frac{R}{T_{rot}})^2  

The tabulated data for the sun are:

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Let's calculate.

           

          K_{sum} = 1.40 \ 10^{36} J

The relationship of the kinetic energy of the sun and the Earth is:

        \frac{K_{sum}}{K_{earth}} = \frac{1.40 \ 10^{36}}{2.66 \ 10^{33}}  

       \frac{K_{sum}}{K_{earth}} =  5.3 \ 10^2  

In conclusion using the definition of kinetic energy we can shorten the result for the relationship between the energy of the sun and the Earth is:

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