Answer to Question #253943 in Physics for Kristine

Question #253943

A body weighing 64𝑙𝑏 slides down from rest at the top of a plane 18𝑓𝑑 long and inclined 30π‘œ above the horizontal. The coefficient of friction is 0.10. Find the velocity of the body as it reaches the bottom of the plane.


1
Expert's answer
2021-10-25T10:03:45-0400

The work-energy theorem gives:


mv22=W1+W2=βˆ’ΞΌmglcos⁑θ+mglsin⁑θ\frac{mv^2}{2}=W_1+W_2=-\mu mgl\cos\theta+mgl\sin\theta

Hence, the speed of a body

v=2gl(βˆ’ΞΌcos⁑θ+sin⁑θ)=2βˆ—32.2βˆ—18βˆ—(βˆ’0.10cos⁑30∘+sin⁑30∘)=22β€…ft/sv=\sqrt{2gl(-\mu \cos\theta+\sin\theta)}\\ =\sqrt{2*32.2*18*(-0.10 \cos30^{\circ}+\sin30^{\circ})}\\ =22\:\rm ft/s


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