Question

A piece of electronic equipment that is surrounded by packing material is dropped so that it hits the ground with a speed of 8.5 m/s. After contact, the equipment experiences an acceleration of a = –kx, where k is a constant and x is the compression of the packing material. If the packing material experiences a maximum compression of 110 mm, at which point the box comes to a complete stop, determine the maximum acceleration of the equipment.

Answer 1

The maximum acceleration of the equipment is -656.81 m/s^2

Step-by-step explanation:

Data

velocity when compression starts,
`v_i = 8.5 m/s`

velocity when compression ends,
`v_f = 0 m/s`

position when compression starts,
`x_i = 0 m`

position when compression ends,
`x_f = 110 mm = 0.11 m`

From aceleration definition

`a = v (dv)/(dx)`

From the problem

`a = -kx`

Combining them and integrating

`v (dv)/(dx) = -kx`

`\int_(v_f)^(v_i) v dv = \int_(x_f)^(x_i) -k x dx`

`(1)/(2) v_f^2 - (1)/(2) v_i^2 = -k * ((1)/(2) x_f^2 - (1)/(2) x_i^2)`

`v_i^2 = k * x_f^2`

`k = (v_i^2)/(x_f^2)`

`k = \frac{{8.5 m/s}^2}{{0.11 m}^2}`

`k = 5971 (1)/(s^2)`

Finally, maximum acceleration is

`a = -kx`

`a = -5971 (1)/(s^2) * 0.11 m`

`a = -656.81 (m)/(s^2)`