yb is therefore vb v298 mjs225 m repeating

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Unformatted text preview: ~skier started from rest, VA == O. Note that the profile of the slope is irrelevant.) Solution If we ignore lossesin energy due to air resistance etc., the mechanical energy of the bow and arrow is the same just before shooting the arrow and when the arrow is at its maximum height, Uo + Ko = U + K. Before shooting, Ko = 0, and Uo = ~kx2 + mgyo, the potential energy of the taut bow plus the gravitational energy of the arrow at the initial position. At the maximum height, K =0 (instantaneously) and U = mgYmax.Therefore, ~kx2 + mgyo = mgYmax,or Ymax -Yo = kx2/2mg = (430 N/m)(0.71 m)2 + 2(0.12kg)(9.8 m/s2) = 92.2 m. (It is assumed that any change in the gravitational potential energy of the bow is negligible.) 8 ... .... ... If 25m -- -1 ;,t 4"em -------- 38~ ProWem 20 Solution >Navy jet of mass 10,000 kg lands on an aircraft ~rier and snags a cable to slow it down. The .llble is attached to a spring with spring constant i~U;OOO NJm. If the spring stretches 25 m to stop plane, what was the landing spee...
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This note was uploaded on 07/27/2009 for the course PHYSICS 101 taught by Professor Wormer during the Spring '08 term at NYU Poly.

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