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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 />

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