Question

A woman steps on a bathroom scale containing a stiff spring. In equilibrium the spring is compressed 1 cm under her weight. The total work done on the spring during the compression is 3.1 J. Find the woman's mass.

Answer 1

To solve this problem, we can utilize the concept of work done by the spring. When an object compresses a spring, the work done is equal to the potential energy stored in the spring, which can be expressed as:

$\[ W = \frac{1}{2} kx^2 \]$

Where:

W = work done (3.1 J)

k = spring constant

x = compression (1 cm = 0.01 m)

We need to find the spring constant. To do this, we'll use the following equation:

\[W = mgh\]

Where:

m = mass of the woman

g = acceleration due to gravity (9.81 m/s^2)

h = height of the compression (0.01 m)

Rearranging the equation to solve for k:

$\[ k=\frac{2W}{x^2}\]$

Then, we can solve for the spring constant:

$\[k=\frac{2*3.1 J}{(0.01 m)^2} = 620 J/m\]

$

Next, we can use the relationship between spring constant and mass to find the woman's mass:

$\[k = \frac{m}{g}\]$

Rearranging for m:

$\[m = k* g = 620 J/m * 9.81 m/s^2 = 6072 N \]

$

So the mass of the woman is approximately 62 kg.