Physics Solution Manual for 1100 and 2101

In applies in both cases because the tracks are

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Unformatted text preview: ance are being ignored. The normal force from the slide is perpendicular to the motion, so it does no work. Thus, no net work is done by nonconservative forces, and the principle of conservation of mechanical energy applies. SOLUTION Applying the principle of conservation of mechanical energy to the swimmer at the top and the bottom of the slide, we have 1 1 2 2 mvf + mghf = mv0 + mgh0 24 14 244 14 244 324 3 Ef E0 If we let h be the height of the bottom of the slide above the water, hf = h , and h0 = H . Since the swimmer starts from rest, v0 = 0 m/s, and the above expression becomes 12 v 2f + gh = gH Solving for H, we obtain v2 H = h+ f 2g Before we can calculate H, we must find vf and h. Since the velocity in the horizontal direction is constant, ∆ x 5.00 m vf = = = 10.0 m/s ∆ t 0.500 s Chapter 6 Problems 311 The vertical displacement of the swimmer after leaving the slide is, from Equation 3.5b (with down being negative), 1 2 2 1 2 y = 2 ay t = 2 (–9.80 m/s )(0.500 s) = −1.23 m Therefore, h = 1.23 m. Using these values of vf and h in the above expression for H, we find v2 (10.0 m/s)2 H = h + f = 1.23 m + = 6.33 m 2g 2(9.80 m/s2 ) ______________________________________________________________________________ 46. REASONING We are neglecting air resistance and friction, which eliminates two nonconservative forces from consideration. Another nonconservative force acting on the semitrailer is the normal force exerted by the ramp. But this force does no work because it is perpendicular to the ramp and the semitrailer’s velocity. Hence, the semi-trailer’s total mechanical energy is conserved, and we will apply the energy conservation principle to determine the initial velocity v0 of the truck. SOLUTION The truck coasts to a stop, so we know that vf = 0 m/s. Therefore, the energy conservation principle gives 1 mv 2 + mgh 1 mv 2 + mgh (144244f ) = (1442440 ) f 0 2 32 3 Ef or m g ( hf − h0 ) = 1 2 2 2 m v0 or g ( hf − h0 ) = 1 v0 (1) 2 E0 Note that hf − h0 is the vertical height up which the truck climbs on the ramp, as the drawing indicates. The drawing also indicates that hf − h0 is related to the distance d the truck coasts along the ramp and the ramp’s angle θ of inclination. The side hf − h0 is opposite the angle θ, and d is the hypotenuse, so we solve sin θ = d hf−h0 θ hf − h0 (Equation 1.1) to find that hf − h0 = d sin θ . d Substituting d sin θ for hf − h0 in Equation (1), we obtain 2 gd sin θ = 1 v0 2 or 2 v0 = 2 gd sin θ or v0 = 2 gd sin θ When the semitrailer enters the ramp, therefore, its speed is ( ) v0 = 2 9.80 m/s 2 (154 m ) sin14.0o = 27.0 m/s 312 WORK AND ENERGY 47. REASONING AND SOLUTION If air resistance is ignored, the only nonconservative force that acts on the skier is the normal force exerted on the skier by the snow. Since this force is always perpendicular to the direction of the displacement, the work done by the normal force is zero. We can conclude, therefore, that mechanical energy is conserved. 1 2 mv0 2 1 2 + mgh0 = mvf2 + mghf Since the skier starts from rest v0 = 0 m/s. Let hf define the zero level for heights, then the final gravitational potential energy is zero. This gives 1 2 mgh0 = mvf2 (1) At the crest of the second hill, the two forces that act on the skier are the normal force and the weight of the skier. The resultant of these two forces provides the necessary centripetal force to keep the skier moving along the circular arc of the hill. When the skier just loses contact with the snow, the normal force is zero and the weight of the skier must provide the necessary centripetal force. mg = mvf2 so that r vf2 = gr F N mg (2) Substituting this expression for vf2 into Equation (1) gives 1 2 mgh0 = mgr Solving for h0 gives r 36 m = = 18 m 2 2 ______________________________________________________________________________ h0 = 48. REASONING The principle of conservation of mechanical energy provides the solution to both parts of this problem. In applies in both cases because the tracks are frictionless (and we assume that air resistance is negligible). When the block on the longer track reaches its maximum height, its final total mechanical energy consists only of gravitational potential energy, because it comes to a momentary halt at this point. Thus, its final speed is zero, and, as a result, its kinetic energy is also zero. In contrast, when the block on the shorter track reaches the top of its trajectory after leaving the track, its final total mechanical energy consists of part kinetic and part potential energy. At the top of the trajectory the block is still moving horizontally, with a velocity that equals the horizontal velocity component that it had when it left the track (assuming that air Chapter 6 Problems 313 resistance is negligible). Thus, at the top of the trajectory the block has some kinetic energy. We can expect that the height H is greater than the height H1 + H2. To see why, note that both blocks have the same initial kinetic energy, since each has the same initial speed. Moreover, the total mechanical e...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.

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