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Wize High School Grade 12 Physics Textbook > Work and Energy

Conservation of Energy

Conservative and Non-Conservative ForcesPrevious Section
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Energy Conservation



The total mechanical energy of a system is the sum of kinetic and potential energies at a given time.

Remember that Wnet=ΔKW_{net}=\Delta Kusing Work-Energy Theorem. We also remember that for conservative force, Wc=−ΔUW_c=-\Delta U. Hence we have:



ΔK=Wnet=Wc+Wnc=−ΔU+Wnc\Delta K=W_{net}=W_c+W_{nc}=-\Delta U+W_{nc}



Solving for WncW_{nc} we will find:


Wnc= ΔK + ΔU=ΔE\boxed{W_{nc}=\ \Delta K\ +\ \Delta U=\Delta E}

where WncW_{nc} is the total work done by non-conservative forces and ΔE\Delta E is the total change in the mechanical energy

Watch Out!
WncW_{nc} could be positive, negative or zero:
  • If it is positive, the mechanical energy is increased (such as pulling force)
  • If it is negative, the mechanical energy is decreased (such as friction)
  • if it is zero. energy is conserved

The above equation sometimes is written is the following form:



Wadded+Ki+Ui=Kf+Uf+WlostW_{added}+K_{i}+U_{i}=K_{f}+U_{f}+W_{lost}



where iiand ffindexes refer to initial and final situations respectively. Furthermore, here WaddedW_{added} and WlostW_{lost} are positive and negative non-conservative works respectively.


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Energy Conservation

If there is no non-conservative work, and there is no dissipated energy, there is no change in energy:

ΔE = 0  →  ΔK +ΔU=0\Delta E\ =\ 0\ \ \rightarrow \ \ \Delta K\ + \Delta U=0

- This is the conservation of energy: energy is not destroyed or created, but transformed into different forms.


Exam Tip
We can use conservation of energy to solve dynamics problem when we are interested in knowing initial and final conditions, without being too concerned about what happens exactly in between.









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Example: Work Done by Friction


A car parked on a 32- degree steep hill loses its traction and starts to slide down. It travels 160 meters down the hill until it reaches speed of 72 km/h. What is the kinetic coefficient of friction between the car and the surface?


Solution:

The car at the top is at rest, while at the bottom, it reaches the velocity of 72 km/h (20 m/s). The only non conservative force that does work on the car is the friction of the surface which does a negative work. So, the mechanical energy is not conserved.

Thus:
Wnc=Ef−EiW_{nc}=E_f-E_i

Wfric=(K+Ug)f−(K+Ug)iW_{fric}=\left(K+U_g\right)_f-\left(K+U_g\right)_i

fk×d×(−1)=(K+0)f−(0+Ug)if_k\times d\times\left(-1\right)=\left(K+0\right)_f-\left(0+U_g\right)_i


Here we set the zero of gravitational potential energy at the end of the path. Note the negative sign of friction work due to cos⁡(180°)\cos(180\degree)
Furthermore, kinetic friction can be written as: fk=μkN=μkmgcos⁡(32)f_k=\mu_kN=\mu_k mg\cos(32)

μk(mg×cos⁡(32))×d×(−1)=12mvf2−mghi\mu_k\left(mg\times\cos\left(32\right)\right)\times d\times\left(-1\right)=\frac{1}{2}mv_f^2-mgh_i
This is a very important step. What we know is that the car has travelled d meters down the hill. But to find hih_iwe need to find the vertical change in the position using a little bit of geometry:
μk(mg×cos⁡(32))×d×(−1)=12mvf2−mg(d sin⁡(32))\mu_k\left(mg\times\cos\left(32\right)\right)\times d\times\left(-1\right)=\frac{1}{2}mv_f^2-mg\left(d\ \sin\left(32\right)\right)
Mass is cancelled from both sides of equations. Thus:
μk(g×cos⁡(32))×d×(−1)=12(20)2−g(d sin⁡(32))\mu_k\left(g\times\cos\left(32\right)\right)\times d\times\left(-1\right)=\frac{1}{2}\left(20\right)_{ }^2-g\left(d\ \sin\left(32\right)\right)

μk(9.81×cos⁡(32))×(160)×(−1)=12(400)−(9.81)(160)(sin⁡(32))\mu_k\left(9.81\times\cos\left(32\right)\right)\times\left(160\right)\times\left(-1\right)=\frac{1}{2}\left(400\right)-\left(9.81\right)\left(160\right)\left(\sin\left(32\right)\right)

μk=0.47\mu_k=0.47




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A 2.00-kg block is pushed against a spring with negligible mass and force constant k= 400 N/m, compressing it 0.220 m. When the block is released , it moves along a friction-less , horizontal surface and then up a friction-less incline with slope 37.0∘37.0^{\circ}

(a) What is the speed of the block as it slides along the horizontal surface after having lost the contact with the spring?
(b) How far does the block travel up the incline before starting to slide back down?




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Example: Rollercoaster (Conservation of Energy)

A rollercoaster starts off at rest at the top of a track. Rank the points along the track from slowest to fastest. Assume there is no drag or friction.




Because there is no drag or friction, we only have one force working on the cart, gravity! So there is no nonconservative forces. This means that we have Wnc = 0 = ΔEW_{nc}\ =\ 0\ =\ \Delta E .

ΔE = ΔKE + ΔPE =0   →    ΔKE = −ΔPE = Wgravity\Delta E\ =\ \Delta KE\ +\ \Delta PE\ =0\ \ \ \rightarrow\ \ \ \ \Delta KE\ =\ -\Delta PE\ =\ W_{gravity}

Basically, all work is done by gravity. So we expect the greatest LOSS change in potential energy is going to give us the greatest GAIN in kinetic energy (or speed). Since gravitational potential is PEg = mghPE_g\ =\ mgh , the largest drop in height will be the point of highest speed.

The ranking is A < D < B < E < C.





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Example: Conservation of Energy on Spring, Mass, Incline System

A 1.20 kg block is at rest on a spring at the base of a 35º degree inclined plane. If the spring constant is k=105 N/m and the mass compresses the spring by 12.0 cm, how far will the block travel up the incline when it is released? (No Friction)
Solution:




m=1.2 kg
K=105 N/m
Δxo=12 cm=0.12m\Delta x_o=12\ cm=0.12m

(i) vi=0⇒\Rightarrow (ii) vf=0⇒\Rightarrow
K⋅E=12mvi2=0K\cdot E=\frac{1}{2}mv_i^2=0 K.E=0, PE=mgh
P.E=12kΔxo2P.E=\frac{1}{2}k\Delta x_o^2

Conservation of energy ; 12kΔxo2=mgh          h=kΔxo22mg=0.064m\frac{1}{2}k\Delta x_o^2=mgh\ \ \ \ \ \ \ \ \ \ h=\frac{k\Delta x_o^2}{2mg}=0.064m
h = 0.064 mh\ =\ 0.064\ m
sin⁡ 35°=hx                     x=hsin⁡35° =11cm\sin\ 35\degree^=\frac{h}{x}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x=\frac{h}{\sin35^{\degree}}\ =11cm
x=11 cmx=11\ cm

Practice: Graphical Problem of Conservation of Energy

A 1.2-g particle is moving along x-axis under the influence of the following potential energy. The particle turns around at x=6mx=6m. What is the maximum speed of this particle?