**PHY 1151**

**Chapter 8; Potential Energy &
Conservative Forces**

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Chapter 8 | Course Calendar
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PE = m g h = (70 kg)(9.8 m/s

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k = = = 167

PE

PE

PE

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a) With what speed does the arrow leave the bow?

b) How high does the arrow go?

c) What average upward force does the ground exert on the arrow while stopping it?

a)

E_{f}= (^{1}/_{2}) m v^{2}= E_{i}= W = (75 N)(0.6 m) = 45.00 J

v

v = 30 m/s

b)E_{f}= m g h = 45 J

(0.1 kg) (9.8 m/s^{2}) h = 45 J

h = 45.9 m

c)W = F D = 45.0 J

F (0.2 m) = 45.0 J

F = 225 N

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a) Find its speed at point B, the bottom of the swing.

A peg is fixed at point O, a distance y directly below point P. The cord hits the peg and the ball swings upward in an arc as shown.

b) Find the ball's speed at point C when the cord between it and the peg is horizontal.

c) Find the ball's height (relative to its starting position) at point D, the top of its swing after hitting the peg (Assume y > L/2).

d) Find the ball's speed at point D, the top of its swing after hitting the peg.

a)At position A, v_{A}= 0 or KE_{A}= 0

If we measure all the vertical distances from y_{B}= 0 at position B, then at A, the mass has a height y_{A}= L, so thatPE_{A}= m g y_{A}= m g L

( This choice for y_{B}= 0 means PE_{B}= 0 ).Therefore, the total energy at A, E

_{A}, isE_{A}= KE_{A}+ PE_{A}= 0 + m g L = m g LEnergy is conserved so

E_{B}= E_{A}

KE_{B}= (^{1}/_{2}) m v_{B}^{2}

PE_{B}= m g y_{B}= 0Remember, we have measured vertical distances here so we have set y

_{B}= 0. That means our total energy at position B isE_{B}= KE_{B}+ PE_{B}= (^{1}/_{2}) m v_{B}^{2}+ 0 = (^{1}/_{2}) m v_{B}^{2}

E_{B}= (^{1}/_{2}) m v_{B}^{2}= m g L = E_{A}

v_{B}^{2}= 2 g L

v_{B}=A peg is fixed at point O, a distance y directly below point P. The cord hits the peg and the ball swings upward in an arc as shown.

At position C, the (vertical) height above position B is

c)y_{C}= (L - y)Energy is conserved so

E_{C}= E_{A}

KE_{C}= m v_{C}^{2}

PE_{C}= m g y_{C}= m g (L - y)That means our total energy at position C is

E_{C}= KE_{C}+ PE_{C}= (^{1}/_{2}) m v_{C}^{2}+ m g (L - y)

E_{C}= (^{1}/_{2}) m v_{C}^{2}+ m g (L - y) = m g L = E_{A}

(^{1}/_{2}) m v_{C}^{2}+ m g L - m g y = m g L

(^{1}/_{2}) m v_{C}^{2}= m g y

v_{C}^{2}= 2 g y

v_{C}=

d)At position D, the (vertical) height above position B isy_{D}= 2 (L - y)as can be seen by taking a more careful look at the diagram.

Energy is conserved soE_{D}= E_{A}

KE_{D}= (^{1}/_{2}) m v_{D}^{2}

PE_{D}= m g y_{D}= m g [2 (L - y)]That means our total energy at position D is

E_{D}= KE_{C}+ PE_{C}= (^{1}/_{2}) m v_{C}^{2}+ m g [2 (L - y)]

E_{D}= (^{1}/_{2}) m v_{D}^{2}+ m g [ 2 (L - y) ] = m g L = E_{A}

(^{1}/_{2}) m v_{D}^{2}+ 2 m g L - 2 m g y = m g L

(^{1}/_{2}) m v_{D}^{2}= 2 m g y - m g L

v_{D}^{2}= 2 g ( 2 y - L )

v_{D}=

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**| ToC,
Chapter 8 | Course Calendar
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**(c) 2005, Doug Davis; all rights reserved.**