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A 29.0-kg block is initially at rest on a horizontal surface. A horizontal force

A 29.0-kg block is initially at rest on a horizontal surface. A horizontal force of 71.0 N is required to set the block in motion, after which a horizontal force of 56.0 N is required to keep the block moving with constant speed.
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catt707p38uu0
catt707p38uu0
4,8(90 marks)

(a) \mu_s=0.25

(b) \mu_k=0.20

Explanation:

According to Newton's second law:

\sum F_y:N=mg\\\sum F_x:F_a=F_f

Recall that the frictional force is related jointly with the coefficient of friction and normal force F_f=\mu N. Replacing in the above equation, we get the coefficient of friction:

F_a=\mu N=\mu mg\\\mu=\frac{F_a}{mg}

(a) The  coefficient of static friction is related with the force required to set the block in motion:

\mu_s=\frac{71N}{29kg*9.8\frac{m}{s^2}}\\\mu_s=0.25

(b) The  coefficient of kinetic friction is related with the force required to keep the block moving with constant speed:

\mu_k=\frac{56N}{29kg*9.8\frac{m}{s^2}}\\\mu_k=0.20

brooklyn4932
brooklyn4932
4,5(53 marks)

i ate apple pie yesterday

Explanation:

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