Because the raindrops do not rebound, the KE (kinetic energy) of each drop is dissipated upon impact.
Given:
m = 1.6 mg = 1.6 x 10⁻³ kg per drop
v = 25 m/s, the speed of each drop
4 drops fall per second.
By definition,
KE = (1/2)*m*v²
Therefore energy dissipated per second is
[tex]E= \frac{1}{2}(1.6 \times 10^{-3} \, \frac{kg}{drop} )*(4 \, \frac{drops}{s}) *(25 \, \frac{m}{s} )^{2} = 2 \, \frac{J}{s} [/tex]
Another way to look at the problem is in terms of momentum of impact.
Because there are 4 drops per second, the time per impact may be approximated as Δt = 1/4 = 0.25 s.
The momentum is
P = F*Δt = m*v
where F = the force of impact
Therefore
F = (m*v)/Δt
= [(1.6 x 10⁻³ kg)*(25 m/s)] / (0.25 s)
= 0.16 N
Answer: 0.16 N
