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## Homework Statement

[PLAIN]http://img830.imageshack.us/img830/3263/19000013.jpg [Broken]

A 12kg box initially at rest slides from the top of an incline. If the force of friction along the incline is 5.0N, what is the speed of the object as it reaches the bottom of the incline?

My book is contradicting itself?

Textbook said:ΔKE + ΔPE + ΔTE = 0

ΔKE + ΔPE = -ΔTE

½m(v² - v₀²) + mg(h - h₀) = -F_{fr}x

½(12kg)(v² - 0) + (12kg)(9.8m/s^{2})(0 - 6.0m) = -(5.0N)(11.0m)

v = [tex]\sqrt{\frac{651J}{6.0kg}}[/tex]

v = 10m/s

Here is what I think it should be

ΔKE + ΔPE + ΔTE = 0

½m(v² - v₀²) + mg(h - h₀) + Ffr(x - x₀) = 0

½(12kg)(v² - 0) + 12(9.8)(0 - 6) + (5.0N)(0 - 11m) = 0

6v² - 705.6J - 55J = 0

v = [tex]\sqrt{\frac{760J}{6kg}}[/tex] = 11.25m/s

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