A0.1 kg apple drops 2m hitting a spring. the spring deforms 0.1m.
what is the spring constant ?

392 N/m

Explanation:

We know that the total mechancial enery will be conserved from when the apple drops to when it hit the spring.  Therefore, we can set the potential energy of the apple equal to the elastic potential energy of the spring.

PE = mgh and SE = 1/2kx² — PE is the initial potential energy of the apple and SE is the elastic potential energy of the spring.

mgh = 1/2kx².  (.1)(9.8)(2) = (1/2)(k)(.1²), solve for k. k = 392

r = ρ x length / area  

ρ = r x area / length  

ρ is in ohm-meters  

in basic units, it is m³kg/s³a²  

i think this is it.

not too sure thinking all accelerations are pointed down