1. A mass m slides without friction down a slide and through a loop. If the mass remainis in the track (even at the top of the loop radius r) what hight must it be released from?
2. A 17 kg child descends a slide 3.5 m high and reaches the bottom with a speed of 2.5 m/s. How much thermal energy is released in the process?
2007-12-05 11:41:49 · 3 answers · asked by brubabe 5 in Science & Mathematics ➔ Physics
1. Height must be such that centripetal acceleration v^2/r = g, or v^2 = gr. To have this velocity at height r, the distance H above r must satisfy H = v^2/(2g). Substituting, H = gr/(2g) = r/2. So the mass must be released from r/2 above the top of the loop.
2. The energy change is mv^2/2 - mgh (a negative number). That's the thermal loss.
2007-12-05 13:22:19 · answer #1 · answered by kirchwey 7 · 0⤊ 0⤋
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2007-12-08 15:34:45 · answer #2 · answered by <3 2 · 0⤊ 0⤋
Look at your formulas, Im sure there has to be a site somewere with them.
If not sure put down a reasonable answer even if it is wrong, mabey you will get some credit?
2007-12-05 11:44:32 · answer #3 · answered by Anonymous · 0⤊ 1⤋