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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 semitrailer’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.
 Spring '13
 CHASTAIN
 Physics, The Lottery

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