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S3.7 Energy conservation

[A] Conservation laws in physics

[B] Energy Conservation

Energy cannot magically appear or disappear.

Key Point 3.15

Energy Conservation. In a closed system, energy is conserved; it can only be transformed between one form and another.

This law is useful because it tells us something that does not change, even when various types of energy are changing.

Demos

Demos

(Demonstration of conservation of energy using air track, pendulum etc)
 

Worked example

Worked example

The system shown is released from rest with the 12kg block 3 m above the floor. Use the principle of conservation of energy to find the velocity with which the block strikes the floor. Neglect friction and inertia of the pulley.
Graphic - No title - pulley

Solution

The total mechanical energy is conserved:

Ki+Ui=Kf+Uf
 
\[ K_i + m_1gh^i_1 +m_2gh^i_2 = K_f + m_1gh^f_1 +m_2gh^f_2 \]
Now, Ki=0, $h_1^i=0$, $h_1^f=3m$, $h_2^i=3m$ $h_2^f=0$.

Hence

12kg×10ms-2×3m=Kf+4kg×10ms-2×3m
Giving
Kf=240J
But
Kf=(m1+m2)v2/2
So
\[ v=\sqrt{30}=5.48 ms^{-1} \]
NB. We could have solved this problem using the methods of free body diagrams and equations of motion S2.5. However, invoking conservation of energy makes it much easier!

 
AConversion between kinetic and potential energy

[C] Violation of energy conservation?

Worked example

Worked example

During a frisbee tournament, a 75g frisbee is thrown from a point 1.5 m above the ground with an initial speed of 15m/s. At some point in its flight it has a height of 2.25m and a speed of 12m/s

  • Calculate the work done on the frisbee by its weight in reaching this point.
  • Is mechanical energy conserved? If not, suggest a reason why.

Solution

  • The frisbee rises 0.75m. Its weight is mg=0.075×10=0.75N directed downwards. Gravity is a conservative force so we can ignore the actual path taken and just use the vertical height change. The work done is then W=-mgΔh=-0.5625J
  • The initial mechanical energy is

    \begin{eqnarray*} %%%Was makeeqnarraystar; label was conserveone K_i + U_i &=& \frac{1}{2}mv_i^2 + mgh_i \\ &=& 8.44J +1.13J \\ &=& 9.57J\; (2sf)\\ \end{eqnarray*}

    The final mechanical energy is

    \begin{eqnarray*} %%%Was makeeqnarraystar; label was conservetwo K_f + U_f &=& \frac{1}{2}mv_f^2 + mgh_f \\ &=& 5.4J + 1.69J \\ &=& 7.09J \;(2sf) \\ \end{eqnarray*}

    Thus, for the Earth-frisbee system, mechanical energy is not conserved. The ‘loss’(=2.48J) is, of course, due to air resistance. Adopting instead the earth-frisbee-air system, we see that energy is conserved.

 
TConservation of Energy: The Swinging Pendulum

Learning Resources

Textbook: HRW Chapter 8.5, 8.8
Course Questions:
Self-Test Questions: