Answer:
Solution
Since this is a linear acceleration problem, we can use Newton’s second law to find the force as long as the electron does not approach relativistic speeds (as long as its speed is much less than 3∗10
8
m/s), which is certainly the case for this problem. We know the initial and final velocities, and the distance involved, so from these we can find the acceleration needed to determine the force.
(a) From v
f
2
=v
f
2
+2ax and ∑F=ma, we can solve for the accelertaion and then the force a=
2x
v
f
2
−v
i
2
Substituting to eliminate a,∑F=
2x
m(v
f
2
−v
i
2
)
Substituting the given information,
∑F=
2(0.0500m)
(9.11∗10
−31
kg)[(7.00∗10
5
m/s)
2
−(3.00∗10
5
m/s)
2
]
∑F=3.64∗10
−18
N
(b) The Earth exerts on the electron the force called weight,
F
g
=mg=(9.11∗10
−31
kg)(9.80m/s
2
)=8.93∗10
−30
N
The accelerating force is
4.08∗10
11
times the weight of the electron.
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