Ashy A.

asked • 04/10/21

physics question

You load a 7.6g AAA battery into your new mini penlight flashlight. The spring at the bottom of the battery compartment has a k=89 and you compress it 2.0 cm as you screw on the cap. At camp you drop the flashlight and the top breaks off.

a. What is the velocity of the battery as it leaves the top of the spring? (assume no friction)

Answer: 2.1m/s.

b. How far above the top of the spring will the battery fly when it is released? (assume no air resistance) Answer: 0.24m.

1 Expert Answer

By:

Dr Gulshan S. answered • 04/11/21

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Give units of k ?

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Ashy A.

k = 89N/m
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04/11/21

Dr Gulshan S.

EPE = 1/2 Kx^2 = Kinetic energy = 1/2mv^2 V^2 = sqrt ( Kx^2/m)= 4.68 ( Plug in x in m , m in Kg , K in N/m) Gives V^2 = 4.68 Gives V= 2.16m/s Now Using Vi^2 = 2gH H = Vi^2 /2g = 4.68/(2*9.8) = 0.238 m = 0.24 m
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04/11/21

Dr Gulshan S.

Ask me if you have any question!
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04/12/21

Ashy A.

Thank you so much! this helped a lot :))
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04/13/21

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