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Unformatted text preview: must point in a direction that is
opposite to the direction of the displacement.
b. Since the work W is known from part (a), we can use W = ( F cos θ ) s (Equation 6.1) to
determine the magnitude F of the force. The direction of the force is opposite to the
displacement of the asteroid, so the angle between them is θ = 180°.
SOLUTION
a. The workenergy theorem states that the work done by the force acting on the asteroid is
equal to its final kinetic energy minus its initial kinetic energy:
2 W = 1 mvf −
2
= 1
2 1 mv 2
0
2 ( 4.5 × 10 kg ) (5500 m /s )
4 (6.3)
2 − 1
2 ( 4.5 × 10 kg ) ( 7100 m /s )
4 2 11 = −4.5 × 10 J b. The work done by the force acting on the asteroid is given by W = ( F cos θ ) s
(Equation 6.1). Since the force acts to slow down the asteroid, the force and the
displacement vectors point in opposite directions, so the angle between them is θ = 180°.
The magnitude of the force is 294 WORK AND ENERGY −4.5 × 1011 J
W
5
F=
=
= 2.5 × 10 N
6
s cos θ
1.8 × 10 m cos180° ( ) (6.1) ______________________________________________________________________________
2
21. REASONING AND SOLUTION The work energy theorem, W = 1 mvf2 − 1 mv0 , gives
2
2 vf = 2W
2
+ v0
m where W = Fs cos 180° = − (4.0 × 105 N)(2500 × 103 m) = − 1.0 × 1012 J
Now
vf = ( 2 −1.0 × 1012 J
4 ) + (11 000 m/s )2 = 9 × 103 m/s 5.0 × 10 kg
______________________________________________________________________________ 22. REASONING Since the person has an upward acceleration, there must be a net force
acting in the upward direction. The net force ΣFy is related to the acceleration ay by
Newton’s second law, ΣFy = ma y , where m is the mass of the person. This relation will
allow us to determine the tension in the cable. The work done by the tension and the
person’s weight can be found directly from the definition of work, Equation 6.1.
SOLUTION
a. The freebody diagram at the right shows the two forces that act
on the person. Applying Newton’s second law, we have T +y s T − mg = ma y
1 24
43
ΣFy Solving for the magnitude of the tension in the cable yields mg T = m(ay + g) = (79 kg)(0.70 m/s2 + 9.80 m/s2) = 8.3 × 102 N b. The work done by the tension in the cable is
WT = (T cos θ ) s = (8.3 × 102 N) (cos 0°) (11 m) = 9.1 × 103 J
c. The work done by the person’s weight is (6.1) Chapter 6 Problems WW = ( mg cos θ ) s = (79 kg) ( 9.8 m/s 2 ) (cos 180°) (11 m) = −8.5 × 103 J 295 (6.1) d. The workenergy theorem relates the work done by the two forces to the change in the
kinetic energy of the person. The work done by the two forces is W = WT + WW:
WT + WW = 1 mv 2 − 1 mv 2
14 3 2 f 2 0
24 (6.3) W Solving this equation for the final speed of the person gives
2
vf = v0 + 2
(WT + WW )
m 2
( 9.1 × 103 J − 8.5 × 103 J ) = 4 m / s
79 kg
______________________________________________________________________________
= 23. ( 0 m/s ) 2 + SSM REASONING According to the workenergy theorem, the kinetic energy of the
sled increases in each case because work is done on the sled. The workenergy theorem is
1
2 1
2 2
given by Equation 6.3: W = KE f − KE 0 = mvf2 − mv0 . The work done on the sled is given by Equation 6.1: W = ( F cos θ )s . The work done in each case can, therefore, be
expressed as W1 = (F cos 0°) s = 1
mv2
f
2 1
2 2
− mv0 = ∆KE1 and
1
2 1
2 2
W2 = ( F cos 62°)s = mvf2 − mv0 = ∆KE2 The fractional increase in the kinetic energy of the sled when θ = 0° is
∆ KE1 ( F cos 0 °)s
=
= 0.38
KE 0
KE0 Therefore, Fs = (0.38) KE0 (1) The fractional increase in the kinetic energy of the sled when θ = 62° is
∆ KE2 ( F cos 62 °)s
Fs
=
=
(cos 62 °)
KE 0
KE 0
KE 0 Equation (1) can be used to substitute for Fs in Equation (2). (2) 296 WORK AND ENERGY SOLUTION Combining Equations (1) and (2), we have
∆ KE2
Fs
(0.38) KE 0
=
(cos 62 °) =
(cos 62 °) = (0.38)(cos 62 °) = 0.18
KE 0
KE 0
KE 0 Thus, the sled's kinetic energy would increase by 18 % .
______________________________________________________________________________
24. REASONING The initial speed v0 of the skier can be obtained by applying the work2
energy theorem: W = 1 mvf2 − 1 mv0 (Equation 6.3). This theorem indicates that the initial
2
2 2
kinetic energy 1 mv0 of the skier is related to the skier’s final kinetic energy
2
work W done on the skier by the kinetic frictional force according to
1
2 1 mv 2
f
2 and the 2
mv0 = 1 mvf2 − W
2 Solving for the skier’s initial speed gives v0 = vf2 − 2W
m (1) The work done by the kinetic frictional force is given by W = ( f k cosθ )s (Equation 6.1),
where fk is the magnitude of the kinetic frictional force and s is the magnitude of the skier’s
displacement. Because the kinetic frictional force points opposite to the displacement of the
skier, θ = 180°. According to Equation 4.8, the kinetic frictional force has a magnitude of
f k = µk FN , where µ k is the coefficient of kinetic friction and FN is the magnitude of the
normal force. Thus, the work can be expressed as
W = ( f k cosθ )s = ( µ k FN cos180°)s
Substituting this expression for W into Equation (1), we have that
v0 = vf2 − 2 µ F ( cos180° ) s
2W
= vf2 − k N
m
m (2) Since the skier is sliding on level ground, the magnitude of the normal force is equal to the
weight mg of the...
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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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