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Physics Solution Manual for 1100 and 2101

# Since the person starts from rest v0 0 ms since the

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Unformatted text preview: nergy is conserved. Therefore, the initial kinetic energy is converted to potential energy as each block moves upward. On the longer track, the final total mechanical energy is all potential energy, whereas on the shorter track it is only part potential energy, as discussed previously. Since gravitational potential energy is proportional to height, the final height is greater for the block with the greater potential energy, or the one on the longer track. SOLUTION The two inclines in the problem are as follows: H2 H v0 v0 H1 θ θ Shorter track Longer track a. Longer track: Applying the principle of conservation of mechanical energy for the longer track gives 2 2 1 1 = 2 mvf + mgh 2 mv0 + mgh 14 244f 4 3 14 2440 4 3 Final total mechanical energy Initial total mechanical energy Recognizing that vf = 0 m/s, hf = H, and h0 = 0 m, we find that 1 2 2 m ( 0 m/s ) + mgH = 1 mv0 + mg ( 0 m ) 2 2 H= 2 v0 2g = ( 7.00 m/s ) 2 2 ( 9.80 m/s 2 ) = 2.50 m b. Shorter track: We apply the conservation principle to the motion of the block after it leaves the track, in order to calculate the final height at the top of the trajectory, where hf = H1 + H2 and the final speed is vf. For the initial position at the end of the track we have h0 = H1 and an initial speed of v0T. Thus, it follows that mvf2 + mg ( H1 + H 2 ) = 144424443 1 2 Final total mechanical energy 2 1 mv0T + mgH 2 14 244 1 4 3 Initial total mechanical energy 314 WORK AND ENERGY Solving for H1 + H2 gives H1 + H 2 = 2 v0T + 2 gH1 − vf2 (1) 2g At the top of its trajectory the block is moving horizontally with a velocity that equals the horizontal component of the velocity with which it left the track, that is, vf = v0T cos θ. Thus, a value for v0T is needed and can be obtained by applying the conservation principle to the motion of the block that takes place on the track. For the motion on the track the final speed is the speed v0T at which the block leaves the track, and the final height is hf = H1 = 1.25 m. The initial speed is the given value of v0 = 7.00 m/s, and the initial height is h0 = 0 m. The conservation principle reveals that 2 1 2 mv0T + mgH 14 244 1 4 3 = Final total mechanical energy 2 v0T = v0 − 2 gH1 = ( 7.00 m/s ) 2 2 1 2 mv0 + mg ( 0 m ) 1442444 4 3 Initial total mechanical energy − 2 ( 9.80 m/s 2 ) (1.25 m ) = 4.95 m/s For use in Equation (1) we find, then, that the speed with which the block leaves the track is v0T = 4.95 m/s and the final speed at the top of the trajectory is vf = v0T cos θ = (4.95 m/s) cos θ. In addition, hf = H1 + H2 at the top of the trajectory, and h0 = H1 = 1.25 m for the initial position at the end of the track. Thus, Equation (1) reveals that 2 v0T + 2 gH1 − vf2 H1 + H 2 = 2g = ( 4.95 m/s ) 2 + 2 ( 9.80 m/s 2 ) (1.25 m ) − ( 4.95 m/s ) cos 50.0° 2 ( 9.80 m/s 2 ) 2 = 1.98 m As expected, H1 + H2 is less than H, and the block rises to a greater height on the longer track. ______________________________________________________________________________ 49. REASONING AND SOLUTION When the car is at the top of the track the centripetal force consists of the full weight of the car. 2 mv /r = mg Chapter 6 Problems 315 Applying the conservation of energy between the bottom and the top of the track gives 2 (1/2)mv + mg(2r) = (1/2)mv0 2 Using both of the above equations 2 v0 = 5gr so 2 2 2 r = v0 /(5g) = (4.0 m/s) /[5(9.8 m/s )] = 0.33 m ______________________________________________________________________________ 50. REASONING AND SOLUTION If air resistance is ignored, the only nonconservative force that acts on the person is the normal force exerted on the person by the surface. 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 (1) where the final state pertains to the position where the person leaves the surface. Since the person starts from rest v0 = 0 m/s. Since the radius of the surface is r, r h0 = r, and hf = r cos θf where θf is the angle at which the person leaves the surface. Equation (1) becomes mgr = mvf2 + mg ( r cos θ f ) 1 2 θf r cos θf r (2) F N In general, as the person slides down the surface, the two forces that act on him are the normal force FN and the weight mg. The centripetal force required to keep the person moving in the circular path is the resultant of FN and the radial component of the weight, mg cos θ. mg cos θ When the person leaves the surface, the normal force is zero, and the radial component of the weight provides the centripetal force. mg cos θ f = mvf2 r ⇒ vf2 = gr cos θf (3) Substituting this expression for vf2 into Equation (2) gives mgr = 1 mg ( r cos θf ) + mg ( r cos θf ) 2 θ θ θ mg 316 WORK AND ENERGY Solving for θf gives 2 ______________________________________________________________________________ θf = cos −1 = 48° 3 51. REASONING The work Wnc done by the nonconservative force exerted on the surfer by the wave is given by Equation...
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