Apiece of electronic equipment that is surrounded by packing material is dropped so that it hits the ground with a speed of 6.5 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 70 mm, determine the maximum acceleration of the equipment. (round the final answer to the nearest whole number.)

302 m/s^2
   To rephrase this problem. We want to know the acceleration it takes to change velocity by 6.5 m/s over a distance of 70mm, or 0.070 meters. We can solve this problem a couple of ways, but for this problem, I'm going to redefine it in terms of work which is force over distance. So let's see how much energy the item has
 E = 0.5*M*V^2
 E = M*0.5*(6.5 m/s)^2
 E = M*0.5*42.25 m^2/s^2
 E = M*21.125 m^2/s^2

   Now for the work needed
E = M*F*D
 M*21.125 m^2/s^2 = M*F*0.07m
 21.125 m^2/s^2 = F*0.07m
 301.7857143 m/s^2 = F
   So the object underwent an acceleration of 302 m/s^2