A 1.2 gram feather is released from rest from a height of 2.1 meters. The feather hits the ground at 1.5m/s
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A 1.2 gram feather is released from rest from a height of 2.1 meters. The feather hits the ground at 1.5m/s

[From: ] [author: ] [Date: 11-12-09] [Hit: ]
0233 J which is what your answer is. Maybe the answer key is wrong?......
Calculate how much mechanical energy is lost. Calculate the average force of friction applied to the feather. (answer is .233J and .111N)
I keep on getting .0233J and .111N... I don't know what I'm doing wrong. Thank you for your help :)

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Note that the initial potential energy of the feather is:
PE = mgh
= (0.0012 kg)(9.8 m/s^2)(2.1 m) [note: m must be in kg to get Joules as the unit]
= 0.0247 J.

The final KE is:
KE = (1/2)mv^2
= (1/2)(0.0012 kg)(1.5 m/s)^2
= 0.00135 J.

Subtracting these two will give the energy lost to be 0.0233 J which is what your answer is. Maybe the answer key is wrong?
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