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A(n) 93 kg clock initially at rest on a horizontal floor requires a(n) 617 N horizontal force to set it in motion. After the clock is in motion, a horizontal force of 528 N keeps it moving with a constant velocity. The acceleration of gravity is 9.81 m/s 2 . a) Find µs between the clock and the floor

Respuesta :

Answer:

[tex]\mu_s = 0.676[/tex]

Explanation:

As we know that the force required to move the clock from rest position must be equal to the maximum limiting friction

So we will have

[tex]F = F_f[/tex]

now we know that

[tex]F_f = \mu_s Mg[/tex]

here we will have

[tex]F = 617 N[/tex]

[tex]m = 93 kg[/tex]

now from above formula we will have

[tex]617 = \mu_s (93)(9.81)[/tex]

[tex]\mu_s = 0.676[/tex]

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