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2 years ago

17. A 25 kg block is initially at rest on a rough, horizontal surface. A horizontal force of 75 N is required to set

Physics

1 answer:

lapo4ka [179]2 years ago

4 0

Answer:

0.30581

0.24464

Explanation:

$\mu_s$ = Coefficient of static friction

$\mu_k$ = Coefficient of kinetic friction

$F_f$ = 75 N

$F_k$ = 60 N

Normal force

$F_n=mg\\\Rightarrow F_n=25\times 9.81\\\Rightarrow F_n=245.25\ N$

Frictional force

$F_f=\mu_sF_n\\\Rightarrow \mu_s=\frac{F_f}{F_n}\\\Rightarrow \mu_s=\frac{75}{245.25}\\\Rightarrow \mu_s=0.30581$

The coefficient of static friction is 0.30581

Kinetic force

$F_k=\mu_kF_n\\\Rightarrow \mu_k=\frac{F_k}{F_n}\\\Rightarrow \mu_s=\frac{60}{245.25}\\\Rightarrow \mu_s=0.24464$

The coefficient of kinetic friction is 0.24464

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Part A:
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Substituting,
d = (0 m/s)(3 s) + 0.5(3 m/s²)(3 s)²
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Refer to the diagram shown below.

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4. The frictional force is μN = μmgcos(30) = 0.866μmg.
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