Q6-17P - University Physics with Modern Physics

Question

(I) How much work is required to stop an electron \(\left( {m = 9.11 \times {{10}^{ - 31}}\;{\rm{kg}}} \right)\)  which is moving with a speed of \(1.10 \times {10^6}\;{\rm{m/s}}\)?

Step-by-Step Solution

Answer

The work required to stop the electron is \( - 5.51 \times {10^{ - 19}}\;{\rm{J}}\).

Step 1: Given data

The mass of the electron is \(m = 9.11 \times {10^{ - 31}}\;{\rm{kg}}\).

The initial speed of the electron is \({v_{\rm{o}}} = 1.10 \times {10^6}\;{\rm{m/s}}\).

The final speed of the electron is \(v = 0\) .

Step 2: Calculation of work

Work done is equal to the change in the kinetic energy of an object under motion.

The initial kinetic energy of the electron is \(K = \frac{1}{2}mv_{\rm{o}}^2\) .

The final kinetic energy of the electron is zero as its final speed is zero.

Thus, the change in kinetic energy is:

\(\begin{aligned}\Delta K = 0 - K\\ =  - K\end{aligned}\) 

The work done is:

\(\begin{aligned}W =  - K\\ =   - \frac{1}{2}mv_{\rm{o}}^2\\ =  - \frac{1}{2} \times \left( {9.11 \times {{10}^{ - 31}}\;{\rm{kg}}} \right) \times {\left( {1.10 \times {{10}^6}\;{\rm{m/s}}} \right)^2}\\ =  - 5.51 \times {10^{ - 19}}\;{\rm{J}}\end{aligned}\) 

(The negative value of work indicates that the electron slows down)

Hence, the work done on the electron is \( - 5.51 \times {10^{ - 19}}\;{\rm{J}}\).