Question

A child bounces in a harness suspended from a door frame by three elastic bands. If each elastic band stretches 0.300 m while supporting a 7.15-kg child at rest, what is the force constant for each elastic band? Assume that each spring supports 1/3 of the child's weight.

Answer 1

Answer:

The force constant for each elastic band is 77.93 N/m

Explanation:

Hooke's law of a spring or an elastic band gives the relation between elastic force (Fe) and stretching (x), the magnitude of that force is:

$F_{e}= kx$ (1)

With k, the elastic force constant. The three elastic bands support the child’s weight (W) and maintain him at rest, so by Newton’s second law for one of the elastic bands:

$\sum F=0$ (2)

$\frac{W}{3}-F_{e}=0\Rightarrow F_{e}=\frac{W}{3}$ (3)

Using (1) on (3):

$kx=\frac{W}{3}\Rightarrow k=\frac{mg}{3x}$(4)

$k=\frac{7.15\,kg*9.81\,\frac{m}{s^{2}}}{3*0.300\,m}\simeq\mathbf{77.93\frac{N}{m}}$