Question

A force of 35.0 n is required to start a 6.0-kg box moving across a horizontal concrete floor. (a) what is the coefficient of static friction between the box and the floor? (b) if the 35.0-n force continues, the box accelerates at 0.60 m/s2 . What is the coefficient of kinetic friction?

Answer 1

convert these from calculater & you will get answer

Answer 2

Given :

Force required to start  a 6.0-kg box moving across a horizontal concrete floor is 35 N .

To Find :

(a) what is the coefficient of static friction between the box and the floor .

(b) if the 35.0-n force continues, the box accelerates at 0.60 m/s2 . What is the coefficient of kinetic friction .

Solution :

(a) It is given that 35 N force is required to start a 6.0-kg box moving across a horizontal concrete floor .

Therefore , it must be equal to static frictional force .

$F=\mu_s(mg)\\\mu_s=\dfrac{F}{mg}\\\\\mu_s=\dfrac{35}{6 \times 10}\\\\\mu_s=0.58$

(b) Now ,

$F-\mu_k(mg)=ma\\\\35-\mu_k(mg)=6\times 0.6\\\\\mu_k(mg)=31.4\\\\\mu_k=\dfrac{31.4}{6\times 10}\\\\\mu_k=0.52$

Hence , this is the required solution .

Learn More :

Kinematics

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