Answer to Question #185037 in Physics for Anne

Question

A  143kg object is dropped from a height of 3m above a spring with a constant 11545N/m. How much (in cm) will the spring be compressed?

Answer 1

According to the energy conservation law, the initial gravitational potential energy of the object was converted into the elastic potential energy of the spring:

$$E_p = E_e$$

The gravitational potential energy is:

$$E_p = mg(h + x)$$

where $m = 13kg$ is the mass of the object, $g = 9.8 m/s^2$ is the gravitational acceleration $h = 3m$ is the height above the spring, and $x$ is the spring compression. Here the fully compressed spring is taken as the zero level for gravitational potential energy.

The elastic potential energy is:

$$E_e =\dfrac{kx^2}{2}$$

where $k = 11545N/m$ is the spring constant.

Thus, obtain the following equation:

$$mg(h+x) = \dfrac{kx^2}{2}\\ \dfrac{kx^2}{2} - mgx - mgh = 0$$

Solving the quadratic equation, obtain:

$$x \approx 0.27\space \text{or} \space x \approx -0.25$$

Since we are interested in comression, take only positive value.

Answer. 0.27 m.