Qu1PracticeSolutions

Qu1PracticeSolutions - Practice Problems Solutions Fall...

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Practice Problems Solutions Fall 2006 - Quiz 1 on Wednesday 27 September 1a) F = ma = m dv dt = mv dv dx = m α x α x 2 . F = 2 /x 3 1b) dx dt = α x assuming x n = 0, xdx = αdt Δ( x 2 ) = α Δ t limit as x i 0 gives Δ t = x 2 f 1c) U ( x ) = i Fdx = i 2 x 3 dx = 2 2 x 2 + C Constraint U ( ) = 0 gives C = 0. U ( x ) = 2 2 x 2 2a) μ s N mg , N = ma x a x g/μ s 2b) Friction between blocks is mg if small block is not falling. In y-direction, for large block : N Mg mg = Ma y = 0 N = ( M + m ) g In x-direction, for large block: F μ k N ma x = Ma x F = ( M + m ) a x + μ k ( M + m ) g ( M + m ) g/μ s + μ k ( M + m ) g F ( M + m ) g ( μ k + 1 s ) 2c) μ k N mg = ma y , N = ma x a y = μ k a x g 3) Each “piece” of rope must be lifted to the table, requiring a work of ( dm ) gh where dm is the mass of the rope piece, g is the gravitational acceleration, and h is the height the piece was lifted. This can be rewritten, ( dm ) gh = ( m l dh ) gh . The total work is just the sum of the work for every piece. W = i d 0 mg l hdh = mg l 1 2 h 2 | d 0 = mgd 2 / 2 l 4a) F = kv 3 = ma = m dv dt = m dv dx dx dt = mv dv dx kv 2 m = dv dx I k m dx = I dv dx v 2 dx
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kx m = 1 v + C Requiring v = v 0 at x = x 0 = 0 allows us to fnd C =
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Qu1PracticeSolutions - Practice Problems Solutions Fall...

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