Chapter 7 The Conservation of Energy
Chapter 7 The Conservation of Energy
Chapter 7 The Conservation of Energy
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124<br />
<strong>Chapter</strong> 7<br />
Picture the Problem Let the system<br />
consist <strong>of</strong> Earth and the child. <strong>The</strong>n<br />
Wext = 0. Choose Ug = 0 at the child’s<br />
lowest point as shown in the diagram to<br />
the right. <strong>The</strong>n the child’s initial energy<br />
is entirely kinetic and its energy when<br />
it is at its highest point is entirely<br />
gravitational potential. We can<br />
determine h from conservation <strong>of</strong><br />
mechanical energy and then use<br />
trigonometry to determineθ.<br />
Using the diagram, relate θ to h and<br />
L:<br />
Apply conservation <strong>of</strong> mechanical<br />
energy to the system to obtain:<br />
Substituting for Ki and Ug,f yields:<br />
Substitute for h in equation (1) to<br />
obtain:<br />
Substitute numerical values and<br />
evaluate θ :<br />
g = U<br />
0<br />
θ<br />
L L − h<br />
vi r<br />
−1⎛<br />
L − h ⎞ −1⎛<br />
h ⎞<br />
θ = cos ⎜ ⎟ = cos ⎜1−<br />
⎟ (1)<br />
⎝ L ⎠ ⎝ L ⎠<br />
W<br />
= ΔK<br />
+ ΔU<br />
= 0<br />
ext<br />
or, because Kf = Ug,i = 0,<br />
− K + U<br />
i<br />
g, f<br />
= 0<br />
2<br />
1 2<br />
vi<br />
− 2 mvi + mgh = 0⇒<br />
h =<br />
2g<br />
θ = cos<br />
θ =<br />
=<br />
−1<br />
⎟ 2 ⎛ v ⎞ i<br />
⎜<br />
⎜1−<br />
⎝ 2gL<br />
⎠<br />
⎛ −1⎜<br />
⎜<br />
1−<br />
⎝ 2<br />
26°<br />
h<br />
2 ( 3.4m/s)<br />
( 9.81m/s<br />
)( 6.0m)<br />
cos 2<br />
45 •• A ball at the end <strong>of</strong> a string moves in a vertical circle with constant<br />
mechanical energy E. What is the difference between the tension at the bottom <strong>of</strong><br />
the circle and the tension at the top?<br />
⎞<br />
⎟<br />
⎠