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Dynamics_Part32

Dynamics_Part32 - mgh 1 − 3 2 μ o cot θ = 1 2 mv 2...

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So, the Principle of Work and Energy tells us that T 1 + U 1 2 = T 2 Denoting the crate’s velocity at the bottom of the ramp by v ,wehave T 1 =0 and T 2 = 1 2 mv 2 Thus, we conclude that
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Unformatted text preview: mgh } 1 − 3 2 μ o cot θ ] = 1 2 mv 2 Solving for the velocity yields v = ± 2 gh } 1 − 3 2 μ o cot θ ]...
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