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Wize University Dynamics Textbook (Master) > Work and Energy

Conservation of Mechanical Energy for a Particle

Conservation of Energy for a ParticlePrevious Section
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Conservation of mechanical energy is a simpler form of conservation of energy principles. It deals specifically with conservative forces. In essence:

ΣT1+ΣV1=ΣT2+ΣV2\Sigma T_1 + \Sigma V_1 = \Sigma T_2 + \Sigma V_2

Where 1 and 2 refer to initial and final conditions, and T and V represent kinetic and potential energy respectively. Comparing this equation to the general conservation of energy equation, the change in potential energy (U1-2) corresponds to (V1-V2). We can use this equation for a single particle, or for a system of particles.

We must emphasize that this equation only applied in the absence of non-conservative forces (typically friction).
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A 2 kg ball is attached to the end of a 5 m cable fixed around a bearing at the top - like a pendulum. The speed of the ball when the pendulum is 30° from the vertical is 2 m/s. Determine:

a) The maximum speed of the ball, and at what point in the motion does this occur?
b) The maximum height of the ball, and the velocity at that point?


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m= 2kg
l = 5m

set potential energy reference @ full cable length


a) T1=12(2)(2)2=4T_1=\frac{1}{2}\left(2\right)\left(2\right)^2=4

v1=2(9.81)(5−5cos⁡30)=13.14v_1=2\left(9.81\right)\left(5-5\cos30\right)=13.14

v2 = 0

T2=12(2)vmax⁡2=17.14 → vmax⁡=4.14 msT_2=\frac{1}{2}\left(2\right)v_{\max}^2=17.14\ \rightarrow\ v_{\max}=4.14\ \frac{m}{s}
@ h=0
v2=0  & T2=17.14=2(9.81)(h)→  h=0.874 m   or  30° to verticalv_2=0\ \ \&\ T_2=17.14=2\left(9.81\right)\left(h\right)\rightarrow\ \ h=0.874\ m\ \ \ or\ \ 30\degree\ to\ vertical

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The system shown below was released from rest with block B at point C. In the position shown, a 30 lb force is applied to bring the system back to point C.

What is the velocity of blocks A and B in the position shown (right before the 30 lb force is applied)?


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