an electron of mass 9.11*10^-31 kg has an initial speed of 4.00*10^5 m/s. it travels in a straight line and its speed increases 8.00*10^5 m/s in a distance of 5.40 cm assume acceleration is constant
determine magnitued of the force exerted on the electron
vf^2=vi^2 + 2 a d
solve for a.
Then, yep, force=mass*acceleration
To determine the magnitude of the force exerted on the electron, you can use Newton's second law of motion, which states that force (F) is equal to mass (m) multiplied by acceleration (a):
F = m * a
In this case, acceleration can be calculated using the formula:
a = (vf - vi) / t
Where:
vf = final velocity
vi = initial velocity
t = time
Given:
Mass of the electron (m) = 9.11 * 10^-31 kg
Initial velocity (vi) = 4.00 * 10^5 m/s
Final velocity (vf) = 12.00 * 10^5 m/s (initial velocity + speed increase)
Distance (d) = 5.40 cm = 0.0540 m
To calculate acceleration:
a = (vf - vi) / t
We need to determine the time (t) taken by the electron to cover the given distance. To find time, we can use the formula:
t = d / v_avg
Where:
d = distance
v_avg = average velocity
To find the average velocity (v_avg), we can consider the initial and final velocities:
v_avg = (vf + vi) / 2
Now, let's calculate the time (t) and average velocity (v_avg):
v_avg = (vf + vi) / 2
= (12.00 * 10^5 + 4.00 * 10^5) / 2
= 8.00 * 10^5 m/s
t = d / v_avg
= 0.0540 / 8.00 * 10^5
= 6.75 * 10^-8 s
Now that we have the time (t), we can calculate the acceleration (a):
a = (vf - vi) / t
= (12.00 * 10^5 - 4.00 * 10^5) / (6.75 * 10^-8)
= 7.41 * 10^12 m/s²
Finally, we can calculate the magnitude of the force exerted on the electron:
F = m * a
= 9.11 * 10^-31 kg * 7.41 * 10^12 m/s²
= 6.75 * 10^-18 Newtons
Therefore, the magnitude of the force exerted on the electron is approximately 6.75 * 10^-18 Newtons.