Question

question 8
use the work - energy theorem to find the force required to accelerate an electron ($m = 9.11\times10^{-31}$ kg) moving along the x axis from $4.00\times10^{6}$ m/s to $1.60\times10^{7}$ m/s in a distance of 0.0125 m.
$3.56\times10^{-19}$ n
$8.20\times10^{-17}$ n
$1.64\times10^{-14}$ n
$5.47\times10^{-22}$ n
$8.75\times10^{-15}$ n

Explanation:

Step1: Apply work - energy theorem

The work - energy theorem states that \(W=\Delta K\), where \(W = Fd\) (work done by force \(F\) over distance \(d\)) and \(\Delta K=K_{f}-K_{i}\) (change in kinetic energy). Kinetic energy \(K=\frac{1}{2}mv^{2}\). So, \(Fd=\frac{1}{2}mv_{f}^{2}-\frac{1}{2}mv_{i}^{2}\).

Step2: Solve for force \(F\)

Rearrange the formula \(F=\frac{m(v_{f}^{2}-v_{i}^{2})}{2d}\).
Substitute \(m = 9.11\times10^{-31}\text{ kg}\), \(v_{i}=4.00\times 10^{6}\text{ m/s}\), \(v_{f}=1.60\times 10^{7}\text{ m/s}\), and \(d = 0.0125\text{ m}\) into the formula.

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Answer:

\(8.75\times 10^{-15}\text{ N}\) (the fifth option)