- #1
IKonquer
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A 4.0 kg block is dragged over a rough horizontal surface by a constant force of 20 N. The speed of the block increases from 3.0 m/s to 5.0 m/s in a displacement of 5.0 m. What is
the magnitude of the work done by the friction force during this displacement?
My work was the following:
Ei + W = Ef
(.5)(m)(3)^2 + W = (.5)(m)(5)^2
And as a result, I got W = 50 - 18 = 32 J.
I'm not sure why this answer is wrong. Thanks in advance.
the magnitude of the work done by the friction force during this displacement?
My work was the following:
Ei + W = Ef
(.5)(m)(3)^2 + W = (.5)(m)(5)^2
And as a result, I got W = 50 - 18 = 32 J.
I'm not sure why this answer is wrong. Thanks in advance.