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

Answer:

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

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

$a = v \frac{dv}{dx}$

From the problem

$a = -kx$

Combining them and integrating

$v \frac{dv}{dx} = -kx$

$\int_{v_f}^{v_i} v dv = \int_{x_f}^{x_i} -k x dx$

$\frac{1}{2} v_f^2 - \frac{1}{2} v_i^2 = -k \times (\frac{1}{2} x_f^2 - \frac{1}{2} x_i^2)$

$v_i^2 = k \times x_f^2$

$k = \frac{v_i^2}{x_f^2}$

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

$k = 5971 \frac{1}{s^2}$

Finally, maximum acceleration is

$a = -kx$

$a = -5971 \frac{1}{s^2} \times 0.11 m$

$a = -656.81 \frac{m}{s^2}$