An electron of mass 9.1E-31 kg is released from rest at a distance of 7.1 E-10 m from a fixed electron. The force on the movable electron is given by the function F = 2.3E-28/r^2, where r is the distance between the electrons. Find the maximum velocity (in meters/second) of the moving electron after release.
^ we're supposed to use mechanical energy theorems to solve this problem.
Ive tried integrating the F(x) to find the total work, but our teacher didn't specify the distance the object travels, so its impossible to bound the integral.
Ideas?
Solutions? thanks!
integrate the repulsion force from the initial separation to infinity and you will get a finite energy change. Set that equal to the final maximum kinetic energy.
integrate with lower bound is 7.1 E-10
and upper bound is infinity.
Then use lim as Rmax approaches infinity, replace upper bound.
Then find velocity by:
Work = (1/2) * m * v^2
v^2 = 2 * Work / m
v = sqrt( 2 * Work / m ).
Answer for this problem: 843780
To solve this problem using mechanical energy theorems, we need to consider the conservation of mechanical energy of the system.
First, let's find the potential energy function for this system. The potential energy (U) of the two electrons due to their electrical interaction is given by Coulomb's Law:
U = k * (q1 * q2) / r
Where k is the electrostatic constant (9 × 10^9 N*m^2/C^2), q1 and q2 are the charges of the electrons (which are equal but opposite, so q1 = -q2), and r is the distance between the electrons.
Since the force (F) is the negative derivative of the potential energy (F = -dU/dr), we find:
F = -dU/dr = -(-k * (q1 * q2) / r^2) = k * (q1 * q2) / r^2
Comparing this with the given force function F = 2.3E-28/r^2, we can determine that:
k * (q1 * q2) = 2.3E-28
Given that the charges of electrons are equal in magnitude (q1 = q2 = e), we can substitute the value of k * (q1 * q2) that we derived:
e^2 = (2.3E-28) / k
The mechanical energy (E) of a system is the sum of its kinetic energy (K) and potential energy (U):
E = K + U
Since the electron is initially at rest, its initial kinetic energy is zero (K_i = 0). At the maximum velocity, when the electron is far away from the fixed electron, the potential energy is zero (U_f = 0). Therefore, the mechanical energy is conserved throughout the motion:
E_i = E_f
K_i + U_i = K_f + U_f
0 + U_i = K_f + 0
U_i = K_f
At the initial position, the distance (r_i) between the electrons is given as 7.1E-10 m. Thus, the initial potential energy (U_i) is:
U_i = k * (q1 * q2) / r_i
And the final kinetic energy (K_f) at maximum velocity is:
K_f = U_i = k * (q1 * q2) / r_i
To find the maximum velocity (v_f) at this position, we can equate the kinetic energy to obtain:
1/2 * m * v_f^2 = k * (q1 * q2) / r_i
Rearranging and solving for v_f, we get:
v_f = sqrt((2 * k * (q1 * q2) / (m * r_i))
Plugging in the known values:
k = 9 × 10^9 N*m^2/C^2
q1 = q2 = 1.6 × 10^-19 C (charge of an electron)
m = 9.1 × 10^-31 kg (mass of an electron)
r_i = 7.1 × 10^-10 m
Substituting these values and calculating, we can find the maximum velocity (v_f) of the moving electron after release.