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Unformatted text preview: in Fig. 830. What is the minimum height h at which it can start from rest and still make it around the loop?
I:. Solution
Problem 29 Solution. From the conservation of mechanical energy (no friction) KA +UA == KB+UB,mgYA = !mv~+mgYB, or v~ = 2g(h  2R). The condition that the block stay on the track is v~ ~ gR (see the solution to Problem 68 below), so 2g(h  2R) ~ gR or h ~ 5R/2. Problem
.30, A runaway truck lane heads uphill at 300 to the i horizontal. H a 16,000kg truck goes out of control and enters the lane going 110 km/h, how far along the ramp does it go? Neglect friction. A
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_ olution yi/g >we neglect losses in mechanical energy, AK = Imv~ = AU = mgy = mgsin 300 Therefore, e == 2. . . 2. = (nO m/3.6 s)2/(9.8 m/s ) = 95;3 m. 1.. I FIGURE 830 Problem 32 Solution. 130 CHAPTER 8
Ii Ii Problem 33. Show that the rope in Example 86 dw remain , taut all the way to the top of its smaller loop only if a ~ ~. (Note thatthe maximum r~lease angle is 90 for the rope to be taut On the way down.)
. Solution :1 'I " I! For the rope to be taut at the top of th...
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 Spring '08
 WORMER
 Physics, Conservation Of Energy, Energy, Work

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