Electron Accleration Calculation | Homework Solutions

[soulja101]

Homework Statement

In an electric tube, an electron of mass 9.1*10 to the power of negative 31 kg experiences a net force of magnitude of 8.0*10 to the power of 15 over a distance of 2.0cm.

A)calculate the magnitude of the electron's acccleration
B)assuming it started from rest, how fast would the electron be moving at the end of 2.0cm motion.

Homework Equations

Fnet=ma

The Attempt at a Solution

a=?
m=9.1*10 to the power of negative 31
d=2.0cm
Fnet-ma
a=fnet/m
=8.0*10 to the power of 15/.1*10 to the power of negative 31 kg

I didnt get how to question B)

 

[OlderDan]

Homework Statement

In an electric tube, an electron of mass 9.1*10 to the power of negative 31 kg experiences a net force of magnitude of 8.0*10 to the power of 15 over a distance of 2.0cm.

A)calculate the magnitude of the electron's acccleration
B)assuming it started from rest, how fast would the electron be moving at the end of 2.0cm motion.

Homework Equations

Fnet=ma

The Attempt at a Solution

a=?
m=9.1*10 to the power of negative 31
d=2.0cm
Fnet-ma
a=fnet/m
=8.0*10 to the power of 15/.1*10 to the power of negative 31 kg

I didnt get how to question B)

Look here

https://www.physicsforums.com/showthread.php?t=110015

to find the 1-D kinematics equation that relates the change in velocity to a known acceleration over a known distance.

 

[m2003]

I would like to clarify that the unit for force is Newtons (N) and for mass is kilograms (kg). Therefore, the correct unit for acceleration is m/s^2.

A) To calculate the magnitude of the electron's acceleration, we can use the formula a = Fnet/m. Plugging in the given values, we get:

a = (8.0*10^15 N) / (9.1*10^-31 kg) = 8.8*10^45 m/s^2

B) To find the final velocity of the electron after 2.0cm of motion, we can use the formula v^2 = u^2 + 2as, where u is the initial velocity (in this case, 0 m/s), a is the acceleration calculated in part A, and s is the distance traveled (2.0cm or 0.02m). Solving for v, we get:

v = √(0^2 + 2*(8.8*10^45 m/s^2)*(0.02m)) = 8.4*10^23 m/s

Therefore, the electron would be moving at a speed of 8.4*10^23 m/s at the end of 2.0cm of motion.