Question

A volcanic ash flow is moving across horizontal ground when it encounters a$\mathbf{10}\mathbf{°}$upslope. The front of the flow then travels 920 mup the slope before stopping. Assume that the gases entrapped in the flow lift the flow and thus make the frictional force from the ground negligible; assume also that the mechanical energy of the front of the flow is conserved. What was the initial speed of the front of the flow?

Short answer

The initial speed of the front of the flow is 56 m/s.

Step-by-step solution

Given data:

The distance travelled by the front of the flow, d = 920m

The angle made by the slope with horizontal, $\mathrm{\theta }=10°$

To understand the concept:

The law of conservation of energy states that energy cannot be created or destroyed - only converted from one form of energy to another. This means that the system always has the same amount of energy unless it is added from outside.

Find theinitial speed of the front of the flowusing the law of conservation energy.

Formula:

$U_i+K{E}_i=U_f+K{E}_f$

Here, $U_i$is the initial potential energy, $K{E}_i$is the initial kinetic energy, $U_f$is the final potential energy, and $K{E}_f$is the final kinetic energy.

Calculate the initial speed of the front of the flow:

According to the conservation of energy,

$$\begin{gathered} U_i+K{E}_i=U_f+K{E}_f \\ mgh=\frac{1}{2}m{v}^2 \end{gathered}$$

Since the slope makes $10°$angle with the horizontal,

$$\begin{gathered} mgd\sin \theta =\frac{1}{2}m{v}^2 \\ {v}^2=2gd\sin \theta \\ v=\sqrt{2gd\sin \theta } \\ v=\sqrt{2(9.8\mathrm{m}/{\mathrm{s}}^2)(920\mathrm{m})\mathrm{s}9\mathrm{n}10°} \\ =\sqrt{(18,032{\mathrm{m}}^2/{\mathrm{s}}^2)(0.174)} \\ =\sqrt{3,137.568{\mathrm{m}}^2/{\mathrm{s}}^2} \\ =56\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the initial speed of the front of the flow is 56 m/s