Selected Hw Problems for Ch. 8

Selected Hw Problems for Ch. 8 - 8.9 In the Fig. a runaway...

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8.9 In the Fig. a runaway truck with failed brakes is moving downgrade at 130 km/h just before the driver steers the truck up a frictionless emergency escape ramp with an inclination of θ = 15 o . The truck’s mass is 1.2x10 4 kg . (a) What minimum length L must the ramp have if the truck is to stop (momentarily) along it? Does the minimum length L increase, decrease, or remain the same if (b) the truck’s mass is decreased and (c) its speed is decreased? By conservation of mechanical energy, mech E0 Δ= UK 0 Δ+ fifi UU KK0 −+ −= Take the initial potential energy to be 0 at the bottom of the ramp where the truck has its initial maximum speed. When the truck is stopped at the top of the ramp distance L and height h from the bottom of the ramp, K f = 0, thus, fi UK0 = =
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2 i 1 mgh mv 2 = But, so, hL s i n 2 i 1 gLsin v 2 θ= 2 i v L 2gsin = θ 2 o 2 km 1 hr m 130 ( ) hr 3.6 km s L 257m m 2(9.8 sin15 ) s ⎛⎞ ⎜⎟ ⎝⎠ == b) The formula derived for L is independent of mass so the length remains the same. c) Since L will be shorter if the initial velocity is lower. 2 i Lv
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8.19 In the Figure with the block released from P and moving along the frictionless track, what are the magnitudes of (a) the horizontal component and (b) the vertical component of the net force acting on the block at point Q ? (c) At what height h should the block be released from rest so that it is on the verge of losing contact with the track at the top of the loop? ( On the verge of losing contact means that the normal force on the block from the track has just then become zero.) (d) Graph the magnitude of the normal force on the block at the top of the loop versus initial height h , for the range h = 0 to h = 6R . By conservation of mechanical energy, mech E0 Δ= UK 0 Δ+ fifi UU KK0 −+ −= 2 Q 1 mgR mgh mv 0 0 2 = ( ) 2 Q v2 g h R = The force acting on the object at Q is the objects mass times its acceleration. In the horizontal direction, centripetal acceleration acting at Q along the horizontal, toward the center, keeps the block on its circular trajectory so,
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2 horizontal c v2 g ( h R ) Fm a m m RR == = (a) (b) In the vertical direction the only force is gravity so F vertical = mg (c) At the top of the loop the same rationale as above gives for the speed at the top of the loop, 2 top 1 mg(2R) mgh mv 0 0 2 −+ = ( ) 2 top g h 2 R =− The forces acting on the block at the top of the loop are the normal force from the track, F N , acting downwards & the force of gravity acting downwards so, a = NC g m a −=
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Where a C is toward the center of the track & thus also downwards. We could have chosen the down direction as positive making all the signs positive (as multiplying through by -1 does now) but making down negative makes these directions explicit. ( ) 2 N 2mg h 2R v Fm g m RR += = Using the hint that the normal force goes to zero for the condition of interest we have, ( ) 2mg h mg R = Solving for h, R5 h2 R R 22 =+ = ( ) R2 R =
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8.20 A single conservative force N, where x is in meters, acts on a particle moving along an x axis. The potential energy U associated with this force is assigned a value of 27 J at x = 0 . (a) Write an expression for
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This note was uploaded on 04/22/2008 for the course PHY 2048 taught by Professor Field during the Spring '08 term at University of Florida.

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Selected Hw Problems for Ch. 8 - 8.9 In the Fig. a runaway...

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