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# Mechanical Energy Study Guide (page 3)

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#### Example

The first swing of a pendulum raises the ball at the end of a string 30.0 cm long at an angle of 60.0° with the vertical. Find the potential energy at the top of the swing if the mass of the ball is 0.25 kg.

#### Solution

The data given in the problem is in SI units. List the known data and draw a diagram of the problem (see Figure 7.5).

l = 30.0 cm

α = 60.0°

m = 0.25 kg

PE = ?

Because the problem does not specify, the choice of the ground level is ours. So, we will consider 0 level as the position where the ball starts moving. At that level, the potential energy is also 0: PE0 = 0 J. The final potential energy will be given by the distance with respect to this ground level.

PE = m · g · h

From the diagram in Figure 7.5, we can see that the height of the ball with respect to the ground is lx. So, we need to find x first and then final potential energy (PE).

x = l · cos α

h = ll · cos α = l · (1 – cos · α)

PE = m · g · l · (1 – cos α) = 37J

## Conservation of Mechanical Energy

The forms of energy discussed previously give the mechanical energy of an object and encompass both the energy of motion and the potential energy:

E = KE + PE

And considering the work-energy theorem discussed previously, we have:

W = ΔKE

W = –ΔPE

Then:

ΔKE = –ΔPE

KEKEo = –(PEPEo)

And rearranging the terms so that each side contains only initial or final states, we write:

KEo + PEo = KE + PE

In other words, the initial and final mechanical energies are equal as long as there are no nonconservative forces in the process or if the work they perform is zero. This establishes the principle of conservation of mechanical energy.

#### Example

Consider an object falling freely from a height of 22 m and stopping on a cliff at a distance of 10 m from the starting point. Find out the speed of the object when it lands on the cliff.

#### Solution

All data is given in the SI units. Therefore, list the data and draw a diagram for this example.

ho = 22 m

h = 22 m – 10 m 12 m

Because the object moved 100 m away from where it was released, it must be closer to the earth.

v = ?

And because this is a conservative process, the energy is conserved, and the mechanical energy at the top and at the time the object hits the cliff are equal. And finally, because the object is falling freely, it means that it started with zero speed:

v0 = 0 m/s

m · g · h0 + m · v02/2 = m · g ·: h + m · v2/2

m · g · h0 = m · g · h + m · v2

The mass can be canceled out through the entire expression:

g · h0 = g · h + v2/2

And we need to solve for v:

v2/2 = g · h0g · h

v2 = 2 · g · (h0h)

v2 = 2 · g · (22 – 12)

### Conservation of Mechanical Energy

The total mechanical energy of an object remains constant provided that the net work done by the external nonconservative forces is zero.

Einitial = Efinal

Practice problems of this concept can be found at: Mechanical Energy Practice Questions

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