## Spring Stretch

Consider a 10 kg object stretches a spring 0.25 m. If the object were swapped out with a larger 15 kg one, how many meters would the spring stretch?

##
__
__**Hint**

**Hint**

A spring’s deflection and force are related by:

$$$F=kx $$$

where
$$k$$
is the spring constant and
$$x$$
is the deflection.

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__
__**Hint 2**

**Hint 2**

First solve for the spring constant with the lower weight. Then, use the spring constant to determine the deflection via the same equation.

A spring’s deflection and force are related by:

$$$F=kx $$$

where
$$k$$
is the spring constant and
$$x$$
is the deflection.

Since
$$Force=mass \times a$$
, we need to multiply the mass by acceleration due to gravity for both objects:

$$$F_1=10kg\times 9.8m/s^2=98\:N$$$

$$$F_2=15kg\times 9.8m/s^2=147\:N$$$

The spring constant is the same for both scenarios, so let’s calculate
$$k$$
for the lower weight because there is only one unknown variable:

$$$k=\frac{F_1}{x_1}=\frac{98N}{0.25m}=392\:N/m$$$

Next, let’s solve for the spring stretch with the larger weight using the spring constant we just determined:

$$$x_2=\frac{F_2}{k}=\frac{147N}{392N/m}=0.375\:m$$$

Alternatively, we could have solved this problem more quickly using a proportional relationship since the spring constant and gravity are the same in both scenarios:

$$$\frac{m_1}{x_1}=\frac{k}{g}=\frac{m_2}{x_2}$$$

$$$\frac{10kg}{0.25m}=\frac{15kg}{x_2}$$$

$$$x_2=\frac{15kg(0.25m)}{10kg}=\frac{3.75m}{10}=0.375\:m$$$

0.375 m