Question

Assuming that Albertine's mass is 60.0 kg , for what value of μk, the coefficient of kinetic friction between the chair and the waxed floor, does she just reach the glass without knocking it over? Use g = 9.80 m/s2 for the magnitude of the acceleration due to gravity.

Answer 1

Answer:

The value of coefficient of kinetic friction is 0.102.

Explanation:

Given that,

Mass of Albertine = 60.0 kg

Suppose Albertine finds herself in a very odd contraption. She sits in a reclining chair, in front of a large, compressed spring. The spring is compressed 5.00 m from its equilibrium position, and a glass sits 19.8 m from her outstretched foot.

If the spring constant is 95.0 N/m.

We need to calculate the value of coefficient of kinetic friction

Using formula of work done

$W =\dfrac{1}{2}kx^2$....(I)

The concept belongs to work energy and power

$W=\mu g h$

Put the value into the formula

$\dfrac{1}{2}kx^2=\mu g h$

Put the value into the formula

$\dfrac{1}{2}\times95.0\times5.00^2=\mu\times60.0\times9.80\times19.8$

$\mu=\dfrac{95.0\times5.00^2}{60.0\times9.80\times2\times19.8}$

$\mu=0.102$

Hence, The value of coefficient of kinetic friction is 0.102.

Answer 2

The coefficient of kinetic friction between the chair and the waxed floor will be  $\mu_{k} =0.101$

What will be the  coefficient of kinetic friction between the chair and the waxed floor /

It is given that,

Mass of Albertine, m = 60 kg

It can be assumed, the spring constant of the spring, k = 95 N/m

Compression in the spring, x = 5 m

A glass sits 19.8 m from her outstretched foot, h = 19.8 m

When she just reaches the glass without knocking it over, a force of friction will also act on it. By conservation of energy, we can balance an equation

$\dfrac{1}{2} kx^{2} =\mu_{k} mgh$

putting the values

$\mu_{k} =\dfrac{kx^{2} }{2mgh}$

$\mu_{k} =\dfrac{95\times 5^{2} }{2\times60\times9.8\times19.8}$

$\mu_{k} =0.101$

Thus the coefficient of kinetic friction between the chair and the waxed floor will be  $\mu_{k} =0.101$

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