cq_1_151

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PHY 241

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A rubber band begins exerting a tension force when its length is 8 cm. As it is stretched to a length of 10 cm its tension increases with length, more or less steadily, until at the 10 cm length the tension is 3 Newtons.

Between the 8 cm and 10 cm length, what are the minimum and maximum tensions?

answer/question/discussion: ->->->->->->->->->->->-> scussion:

Minimum tension 0 Max tension 3N

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Assuming that the tension in the rubber band is 100% conservative (which is not actually the case) what is its elastic potential energy at the 10 cm length?

answer/question/discussion: ->->->->->->->->->->->-> scussion:

The average force through this would be 1.5 N. over 2 cm

1.5*2=3Ncm at the 10 cm or .03J

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If all this potential energy is transferred to the kinetic energy of an initially stationary 20 gram domino, what will be the velocity of the domino?

answer/question/discussion: ->->->->->->->->->->->-> scussion:

KE = .5( m)(v)^2

.03J=.5 0.02kgv^2

V=30m/s

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This is not the correct solution to your equation.

Your equation is OK.

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If instead the rubber band is used to 'shoot' the domino straight upward, then how high will it rise?

answer/question/discussion: ->->->->->->->->->->->-> scussion:

v0=30m/s

vf=0

'dt=?

`ds=?

a=-9.8m/s^2

vf^2 = v0^2 + 2 a `ds

0 = 30m/s^2 + 2 -9.8m/s^2 `ds

`ds=4.69m

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This is incorrect.

You should in any case solve this using energy considerations.

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For University Physics students:

Why does it make sense to say that the PE change is equal to the integral of the force vs. position function over an appropriate interval, and what is the appropriate interval?

answer/question/discussion: ->->->->->->->->->->->-> sion:

The interval would be from 8cm to 10cm.

Ke=force*`dt

Force *position would be the area under the curve if graphed.

The integral is basically the area under the curve.

There is probably a more elegant way to put this but this is how I understand it.

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