Answer: a) 31.07 * 10^6 N/C ; b) 11.68 * 10^6 N/C
Explanation: in order to solve this problem we have to use the Coulomb force which is given by:
F=k*q1*q2/r^2
( see attach for details)
Answer:
Explanation:
given,
mass of block = 3 kg
spring constant k = 500 N/m
kinetic friction coefficient µk = 0.6
speed of block = 5 m/s
F = µk N
F = 0.6 x 3 x 9.8
F = 17.64 N
using energy conservation


250 x² + 17.64 x - 37.5 = 0
on solving
x = 0.354 m
graph is attached below
Fnet=m*a
a= Fnet/m
a=(20N[E]+9.0N[E])/ 12.0kg
a=2.4167m/s^2
a ≈ 2.4m/s^2
If the frequency is large then the period is short and if the frequency is small then the period is long