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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 Spring '13
 CHASTAIN
 Physics, Pythagorean Theorem, Force, The Lottery, Right triangle, triangle

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