6. a wooden chair is pushed across a wooden floor. the friction force is 34.5 n. what
is the weight of the chair? what is the mass of the chair? (=0.30)
7. a copper piggy bank is pushed across a steel counter. the frictional force is 12 n
what is the mass of the piggy bank? (u= 0.36)
8. you are dragging a cat by its tail. the u between your cat and the floor is 0.21. if the
frictional force between your cat and the floor is 2.5 n, find the cat's mass.
i

6) m = 11.73 Kg

7) m = 3.4 Kg

8) m = 1.21 Kg

Explanation:

6) Given,

The friction force, f = 34.5 N

The coefficient of friction, μ = 0.30

The friction force is given by the formula,

                             f = μ · η

Where η is the normal force acting on the body

                             η = m x g

Substituting 'η' in the above equation

                              f = μ · m x g

                              m = f / μg

Substituting the given values in the above equation

                              m = 34.5 / (0.3 x 9.8)

                                  = 11.73 Kg

Hence, the mass of the wooden chair, m = 11.73 Kg

2) Given,

The friction force, f = 12 N

The coefficient of friction, μ = 0.36

Substituting the given values

                           m = 12 / (0.36 x 9.8)

                               = 3.4 Kg

Hence, the mass of the piggy bank, m = 3.4 Kg

3) Given,

The friction force, f = 2.5 N

The coefficient of friction, μ = 0.21

Substituting the given values

                           m = 2.5 / (0.21 x 9.8)

                               = 1.21 Kg

Hence, the mass of the cat, m = 1.21 Kg

r=rho l/a where a is the area, l is the length and rho is resistivity so the greater the area the lesser the resistance thats why the earth pin is thicker bcos current follow the shortest path, so less resistive .