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MissTica
3 years ago
6

A 565 N rightward force pulls a large box across the floor with a constant velocity of 0.75 m/s. If the coefficient of friction

is 0.8, what is the weight of the box?
Physics
1 answer:
alukav5142 [94]3 years ago
6 0
The first thing you should know is that the friction force is equal to the coefficient of friction due to normal force.
 Therefore, clearing the normal force we have:
 The friction is 565N.
 (565 / 0.8) = 706.25N. weight.
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Explanation:

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m=\frac{W}{g} =\frac{455}{9.81}=46.38kg

The second step is to draw the free body diagram of the body (see attached image) and use Newton's second law that states that the sum of the forces is equal to mass by acceleration

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X = displacement =6.8m

Solving

Vf=\sqrt{(6.8)(2)(2)+0^2}=27.2m/s

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