95
vf 2 = 2Wnet ⁄ m = 2 * 132 ⁄ 20 m2 ⁄ s2 = 13.2 m2 ⁄ s2
vf = √(13.2) m ⁄ s = 3.6 m ⁄ s
96
ME is the mechanical energy of an object whose kinetic and potential energies are KE and PE respectively.
The expression for the potential energy varies from force to force. Also, sometimes there may be more than one conservative forces involved. On the other hand, the expression for kinetic energy is always mv2 ⁄ 2.
ME = mv2 ⁄ 2 + PE
Example: An object of mass 4 kg has a speed of 20 m ⁄ s when at a point where the object has a potential energy of 50 J. Calculate its mechanical energy.
Solution: m = 4 kg; v = 20 m ⁄ s; PE = 50 J; ME = ?
ME = mv2 ⁄ 2 + PE = (4 * 202 ⁄ 2 + 50) J = 850 J
If all the forces with a non-zero contribution to the net work done acting on an object are conservative, then the net work done on the object is equal to the work done by the conservative forces. According to the work kinetic energy, the net work done on an object is equal to the change of its kinetic energy.
And the work done by conservative forces is equal to the negative of the change in its potential energy.
Therefore, if all the forces acting on an object are conservative the following equation holds:
ΔKE = -ΔPE
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97
But ΔKE = KEf – KEi and ΔPE = PEf – PEi. After substituting these expressions and collecting the initials on one side and the finals on the other side, the following equation is obtained.
KEi + PEi = KEf + PEf
The left hand side of this equation is equal to the initial mechanical energy of the object and the right hand side of this equation is equal to the final mechanical energy of the object. The mechanical energy of the object remains the same. In other words the mechanical energy of the object is conserved. This is a mathematical statement of the principle of conservation of mechanical energy. The principle of conservation of mechanical energy states that if all the forces with none-zero contribution to the work done acting on an object are conservative, then the mechanical energy of the object is conserved.
Example: A 5 kg object was displaced from a location where its potential energy is 20 J to a location where its potential energy is 10 J under the influence of conservative forces only. If it’s initial speed is 2 m ⁄ s,
a) Calculate its initial kinetic energy.
Solution: m = 5 kg; vi = 2 m ⁄ s; KEi = ?
KEi = mvi 2 ⁄ 2 = (5 * 22 ⁄ 2) J = 10 J
b) calculate its initial mechanical energy.
Solution: PEi = 10 J; MEi = ?
MEi = KEi + PEi = (10 + 20) J = 30 J c) Calculate its mechanical energy at its final location.
Solution: Since all the forces acting on the object are conservative, mechanical energy is conserved.
MEf = ? MEf = MEi = 30 J d) calculate its kinetic energy at the end of the displacement.
Solution: PEf = 10 J; KEf = ?
MEf = KEf + PEf
KEf = MEf – PEf = (30 – 10) J = 20 J
98 e) calculate its speed at the end of the displacement.
Solution: vf = ?
KEf = mvf 2 ⁄ 2
vf = √(2 KEf ⁄ m) = √(2 * 20 ⁄ 5) J = 2.8 m ⁄ s