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Unformatted text preview: can be found from Equation 2.9 with v0 = 0 m/s,
since the rock is released from rest. At a height h, the rock has fallen through a distance
2
(20 .0 m) – h , and according to Equation 2.9, v = 2ay = 2 a[(20.0 m) – h]. Therefore, the
kinetic energy at any height h is given by KE = ma[(20.0 m)– h ]. The total energy at any
height h is the sum of the kinetic energy and potential energy at the particular height.
SOLUTION The calculations are performed below for h = 10.0 m . The table that follows
also shows the results for h = 20.0 m and h = 0 m.
2 PE = mgh = (2.00 kg)(9.80 m/s )(10.0 m) = 196 J
KE = ma[ (20.0 m) – h ] = (2.00 kg)(9.80 m/s 2 )(20.0 m – 10.0 m) = 196 J
E = KE + PE = 196 J + 196 J = h (m)
20.0
10.0
0 KE (J)
0
196
392 PE (J)
392
196
0 392 J E (J)
392
392
392 Note that in each case, the value of E is the same, because mechanical energy is conserved.
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38. REASONING
a. and b. Since there is no friction, there are only two forces acting on each box, the
gravitational force (its weight) and the normal force. The normal force, being perpendicular
to the displacement of a box, does no work. The gravitational force does work. However, it
is a conservative force, so the principle of conservation of mechanical energy applies to the
motion of each box, and we can use it to determine the speed in each case. This principle
reveals that the final speed depends only on the initial speed, and the initial and final
heights. The final speed does not depend on the mass of the box or on the steepness of the
slope. Thus, both boxes reach B with the same speed.
c. The kinetic energy is given by the expression KE = 1 mv 2 (Equation 6.2). Both boxes
2
have the same speed v when they reach B. However, the heavier box has the greater mass.
Therefore, the heavier box has the greater kinetic energy. 306 WORK AND ENERGY SOLUTION
a. The conservation of mechanical energy states that
1
2 2 mvB + mghB = 1 mv 2
A
2 + mghA (6.9b) Solving for the speed vB at B gives vB = vA + 2 g ( hA − hB ) =
2 ( 0 m /s )2 + 2 ( 9.80 m / s 2 ) ( 4.5 m − 1.5 m ) = 7.7 m /s b. The speed of the heavier box is the same as that of the lighter box: 7.7 m /s .
c. The ratio of the kinetic energies is ( KEB )heavier box
( KEB )lighter box (
=
( )
) 1 mv 2
B
2
heavier box
2
1 mv
B
2
lighter box = mheavier box
mlighter box = 44 kg
=
11 kg 4.0 ______________________________________________________________________________
39. REASONING We can find the landing speed vf from the final kinetic energy KEf, since
KEf = 1 mvf2 . Furthermore, we are ignoring air resistance, so the conservation of mechanical
2
energy applies. This principle states that the mechanical energy with which the pebble is
initially launched is also the mechanical energy with which it finally strikes the ground.
Mechanical energy is kinetic energy plus gravitational potential energy. With respect to
potential energy we will use ground level as the zero level for measuring heights. The
pebble initially has kinetic and potential energies. When it strikes the ground, however, the
pebble has only kinetic energy, its potential energy having been converted into kinetic
energy. It does not matter how the pebble is launched, because only the vertical height
determines the gravitational potential energy. Thus, in each of the three parts of the problem
the same amount of potential energy is converted into kinetic energy. As a result, the speed
that we will calculate from the final kinetic energy will be the same in parts (a), (b), and (c).
SOLUTION
a. Applying the conservation of mechanical energy in the form of Equation 6.9b, we have
2
2
1
1
2 mvf + mgh = 2 mv0 + mgh0
14 244f
4
3
14 244
4
3
Final mechanical energy
at ground level Initial mechanical energy
at top of building Solving for the final speed gives
2
vf = v0 + 2 g ( h0 − hf ) = (14.0 m/s ) 2 + 2 ( 9.80 m/s 2 ) ( 31.0 m ) − ( 0 m ) = 28.3 m/s Chapter 6 Problems 307 b. The calculation and answer are the same as in part (a).
c. The calculation and answer are the same as in part (a). 40. REASONING The distance h in the drawing in the text is the difference between the
skateboarder’s final and initial heights (measured, for example, with respect to the ground),
or h = hf − h0. The difference in the heights can be determined by using the conservation of
mechanical energy. This conservation law is applicable because nonconservative forces are
negligible, so the work done by them is zero (Wnc = 0 J). Thus, the skateboarder’s final total
mechanical energy Ef is equal to his initial total mechanical energy E0:
1 mv 2 + mgh = 1 mv 2 + mgh
f 244f
02440
2
2
144
3 144
3 Ef (6.9b) E0 Solving Equation 6.9b for hf − h0, we find that
hf − h0 =
1 24
43
h 1 v2
20 − 1 vf2
2
g SOLUTION Using the fact that v0 = 5.4 m/s and vf = 0 m/s (since the skateboarder comes
to a momentary rest), the distance h is h= 1 v2
20 1 ( 5.4
2 − 1 vf2
2 2 m/s ) − 1 ( 0 m/s )
2 2 =
= 1.5 m
g
9.80 m/s 2
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41. REASONING Since air resistance is being...
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 Spring '13
 CHASTAIN
 Physics, The Lottery

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