Use the work-energy theorem to solve each of these problems. You can use Newton’s laws to check your answers. Neglect air resistance in all cases. (A) A branch falls from the top of a 95.0-m- tall redwood tree, starting from rest. How fast is it moving when it reaches the ground? (b) A volcano ejects a boulder directly upward 525 m into the air. How fast was the boulder moving just as it left the volcano?

The given data is listed below as-

• The initial velocity of the branch is, U = 0 m/s
• The height of the redwood tree is h = 95.0 m
• The height reached by the boulder is h = 525 m

When forces act on a particle it undergoes displacement and the particle’s kinetic energy changes by an amount equal to the total work done on the particle by all the forces. Therefore, work done on the particle is given by-

${\mathbf{W}}_{\mathbf{total}}\mathbf{=}{\mathbf{K}}_{\mathbf{2}}\mathbf{-}{\mathbf{K}}_{\mathbf{1}}\mathbf{=}\mathbf{∆}\mathbf{K}$

The work-energy theorem can be applied to all the bodies that can be treated as particles.

The change in kinetic energy is given by

$∆\mathrm{K}=\frac{1}{2}{\mathrm{mV}}^2-\frac{1}{2}{\mathrm{mU}}^2$

Here, m is the mass of the branch, V is the final speed of the falling branch, and U is the initial speed of the branch.

It is given that the branch starts from the rest. Therefore, the initial speed of the branch

U = 0

$$\begin{gathered} ∆\mathrm{K}=\frac{1}{2}{\mathrm{mV}}^2-0 \\ ∆\mathrm{K}=\frac{1}{2}{\mathrm{mV}}^2 \end{gathered}$$

Now, Work done by gravitational force W = mgh

Here, m is the mass of the branch, g is the acceleration due to gravity, and h is the height.

Now, equate $∆\mathrm{K}$and W,

$$\begin{gathered} \frac{1}{2}{\mathrm{mV}}^2=\mathrm{mgh} \\ {\mathrm{V}}^2=2\mathrm{gh} \\ {\mathrm{V}}^2=2\times 9.8\times 95 \\ \mathrm{V}=43.2\mathrm{m}/\mathrm{s} \end{gathered}$$

Thus, the speed of branch will be 43.2 m/s , when it reaches the ground.

At maximum height, the speed of the boulder is V = 0

To calculate the initial speed of U of the boulder, use the work-energy theorem.

From the work-energy theorem,

$$\begin{gathered} \frac{1}{2}\mathrm{m}\left({\mathrm{u}}^2-{\mathrm{v}}^2\right)=\mathrm{mgh} \\ {\mathrm{u}}^2-{\mathrm{v}}^2=2\mathrm{gh} \end{gathered}$$

Now, V=0,

Therefore,

$$\begin{gathered} {\mathrm{u}}^2-{\mathrm{v}}^2=2\mathrm{gh} \\ \mathrm{u}=\sqrt{2\times 9.8\times 525} \\ \mathrm{u}=\sqrt{10290{\mathrm{m}}^2/{\mathrm{s}}^{-2}} \\ \mathrm{u}=101\mathrm{m}/\mathrm{s} \end{gathered}$$

Thus, when the boulder just leaves the volcano, it speed will be 101 m/s