Q18E - University Physics with Modern Physics

Question

Some typical kinetic energies. (a) In the Bohr model of the atom, the ground-state electron in hydrogen has an orbital speed of 2190 km/s. What is its kinetic energy (consult Appendix F) (b) If you drop a 1.0-kg weight (about 2 lb) from a height of 1.0 m, how many joules of kinetic energy will it have when it reaches the ground? (c) Is it reasonable that a 30.0-kg child could run fast enough to have 100 J of kinetic energy?

Step-by-Step Solution

Answer

(a) $2.18\times {10}^{-18} \mathrm{J}$ 

 

(b)  9.8 J 

 

(c) Yes

Step 1: Identification of the given data

The given data is listed below as-

• The orbital speed of ground state electron in Hydrogen is $V=2190 \mathrm{km}/\mathrm{s}=2190\times {10}^3 {\mathrm{ms}}^{‐1}$ .
• The mass of the child is, $m=30.0 \mathrm{kg}$ 
• Kinetic energy, $K=100 \mathrm{J}$  

Step 2: Significance of the kinetic energy

The kinetic energy of a particle equals the amount of work required to accelerate the particle from rest to speed V. Therefore, kinetic energy on the particle is given by-

$\mathbf{K}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{mV}}^{\mathbf{2}}$ 

The kinetic energy is a scalar and it is always positive or zero.

Step 3: Determination of kinetic energy of the electron

(a)

The mass of the electron is $m_e=9.1\times {10}^{-31} \mathrm{kg}$  

The kinetic energy of an object is given by-

$K=\frac{1}{2}{m}_e{V}^2$   

Here, m is the mass of the electron, and V is the speed of the electron.

For, $m_e=9.1\times {10}^{-31} \mathrm{kg}$ and $V=2190\mathrm{km}/\mathrm{s}=2190\times {10}^3 {\mathrm{ms}}^{‐1}$ 

Therefore, the Kinetic energy of the electron is given by-

$$\begin{gathered} K=\frac{1}{2}{m}_e{V}^2 \\ K=\frac{1}{2}\times \left(9.1\times {10}^{-31} \mathrm{kg}\right)\times {\left(2190\times {10}^3 \mathrm{m}/\mathrm{s}\right)}^2 \\ K=2.18\times {10}^{-18} \mathrm{J} \end{gathered}$$ 

 

Thus, the Kinetic energy of the electron is $2.18\times {10}^{-18} \mathrm{J}$  

Step 4: Determination of kinetic energy when a weight of 1 kg is dropped from a height of 1 m

(b)

The kinetic energy under free fall along the y-axis is given by-

$K=\frac{1}{2}m{V_y}^2$ 

 The velocity at time t of a particle is given by-

$V^2={V_0}^2+2as$  ………..(1)

Now, the object is falling along the y-axis with constant gravitational acceleration and $S=y_2-y_1$  

Therefore, equation (1) becomes

${V_y}^2={V_{0y}}^2-2g\left(y_2-y_1\right)$ 

For, $V_{0y}=0, g=9.8 {\mathrm{ms}}^{‐2}$ and $S=1m$ 

${V_y}^2={V_{0y}}^2-2g\left(y_2-y_1\right)$ will become

 $$\begin{gathered} {V_y}^2=0-2\times 9.8 {\mathrm{ms}}^{‐2}\times 1\mathrm{m} \\ =19.6 {\mathrm{m}}^2{\mathrm{s}}^{‐2} \end{gathered}$$

Therefore, kinetic energy is given by

$$\begin{gathered} K=\frac{1}{2}\times 1\mathrm{kg}\times 19.6{\mathrm{m}}^2{\mathrm{s}}^{‐2} \\ =9.8 \mathrm{J} \end{gathered}$$ 

Thus, kinetic energy when a weight of 1 kg is dropped from a height of 1 m is $9.8 \mathrm{J}$.

Step 5: Determine whether it is reasonable that a 30.0-kg child could run fast enough to have 100 J of kinetic energy

(c)

The mass of the child is $m=30 \mathrm{kg}$ and 

Kinetic energy is $K=100 \mathrm{J}$  

 

$$\begin{gathered} K=\frac{1}{2}m{V_y}^2 \\ 100 J=\frac{1}{2}\times 30 \mathrm{kg}\times {{\mathrm{V}}_{\mathrm{y}}}^2 \\ {V}_y=2.58 {\mathrm{ms}}^{-1} \end{gathered}$$ 

Thus, it is reasonable that a 30.0-kg child could run fast enough to have 100 J of kinetic energy.