ch08-p064 - d replaced by L and the final v 2 term being...

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Thus, we see that its kinetic energy right at the beginning of its “rough slide” (heading uphill towards B ) is K C = 1 2 m (4.980 m/s) 2 = 12.4 m (with SI units understood). Note that we “carry along” the mass (as if it were a known quantity); as we will see, it will cancel out, shortly. Using Eq. 8-37 (and Eq. 6-2 with F N = mg cos θ ) and sin yd θ , we note that if d < L (the block does not reach point B), this kinetic energy will turn entirely into thermal (and potential) energy K C = mgy + f k d ¡ 12.4 m = mgd sin θ μ k mgd cos θ With μ = 0.40 and θ d = 1.49 m, which is greater than L (given in the problem as 0.75 m), so our assumption that d < L is incorrect. What is its kinetic energy as it reaches point B ? The calculation is similar to the above, but with d
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Unformatted text preview: d replaced by L and the final v 2 term being the unknown (instead of assumed zero): 1 2 m v 2 = K C ( mgL sin + k mgL cos ) . This determines the speed with which it arrives at point B : 2 2 2 2 (sin cos ) (4.98 m/s) 2(9.80 m/s )(0.75 m)(sin30 0.4cos30 ) 3.5 m/s. B C k v v g L = + = + = 64. We will refer to the point where it first encounters the rough region as point C (this is the point at a height h above the reference level). From Eq. 8-17, we find the speed it has at point C to be v C = v A 2 2 gh = (8.0) 2 2(9.8)(2.0) = 4.980 5.0 m/s....
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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