8.3 Conservation of energy  (Page 3/9)

Since the particle is released from rest, the initial kinetic energy is zero. At the lowest point, we define the gravitational potential energy to be zero. Therefore our conservation of energy formula reduces to

The vertical height of the particle is not given directly in the problem. This can be solved for by using trigonometry and two givens: the length of the pendulum and the angle through which the particle is vertically pulled up. Looking at the diagram, the vertical dashed line is the length of the pendulum string. The vertical height is labeled h . The other partial length of the vertical string can be calculated with trigonometry. That piece is solved for by

Therefore, by looking at the two parts of the string, we can solve for the height h ,

We substitute this height into the previous expression solved for speed to calculate our result:

Significance

Air resistance on a falling object

A helicopter is hovering at an altitude of $1\text{km}$ when a panel from its underside breaks loose and plummets to the ground ( [link] ). The mass of the panel is $15\text{kg},$ and it hits the ground with a speed of $45\text{m}\text{/}\text{s}$ . How much mechanical energy was dissipated by air resistance during the panel’s descent?

Strategy

Step 1: Here only one body is being investigated.

Step 2: Gravitational force is acting on the panel, as well as air resistance, which is stated in the problem.

Step 3: Gravitational force is conservative; however, the non-conservative force of air resistance does negative work on the falling panel, so we can use the conservation of mechanical energy, in the form expressed by [link] , to find the energy dissipated. This energy is the magnitude of the work:

$\text{Δ}{E}_{\text{diss}}=\left|W_{\text{nc,if}}\right|=\left|\text{Δ}{\left(K+U\right)}_{\text{if}}\right|.$

Step 4: The initial kinetic energy, at $y_{\text{i}}=1\text{km},$ is zero. We set the gravitational potential energy to zero at ground level out of convenience.

Step 5: The non-conservative work is set equal to the energies to solve for the work dissipated by air resistance.

Solution

The mechanical energy dissipated by air resistance is the algebraic sum of the gain in the kinetic energy and loss in potential energy. Therefore the calculation of this energy is

$\begin{array}{cc}\hfill \text{Δ}{E}_{\text{diss}}& =\left|K_{\text{f}}-K_{\text{i}}+U_{\text{f}}-U_{\text{i}}\right|\hfill \\ & =\left|\frac{1}{2}\left(15\text{kg}\right){\left(45\text{m/s}\right)}^2-0+0-\left(15\text{kg}\right)\left(9.8{\text{m/s}}^2\right)\left(1000\text{m}\right)\right|=130\text{kJ}.\hfill \end{array}$

Significance

Most of the initial mechanical energy of the panel $\left(U_{\text{i}}\right)$ , 147 kJ, was lost to air resistance. Notice that we were able to calculate the energy dissipated without knowing what the force of air resistance was, only that it was dissipative.

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