Ski jumping in Vancouver The 2010 Olympic ski jumping competition was held at Whistler Mountain near Vancouver During a jump, a skier starts near the top of the in-run, the part down which the skier glides at increasing speed before the jump. The Whistler in-run is 116 m long and for the first part is tilted down at about 35 ° below the horizontal There is then a curve that transitions into a takeoff ramp, which is tilted 11 ° below the horizontal. The skier flies off this ramp at high speed body tilted forward and skis separated ( Figure 4.15 ). This position exposes a large surface area to the air, which creates lift, extends the time of the jump, and allows the jumper to travel farther In addition, the skier pushes off the exit ramp of the in-run to get a vertical component of velocity when leaving the ramp. The skier lands 125 m or more from the end of the in-run. The landing surface has a complex shape and is tilted down at about 35 ° below the horizontal. The skier moves surprisingly close (2 to 6 m) above the snowy surface for most of the jump. The coefficient of kinetic friction between the skis and the snow on the in-run is about 0.05 ± 0.02 , and skiers’ masses are normally small—about 60 kg. We can make some rough estimates about an idealized ski jump with an average in-run inclination of ( 35 ° + 11 ° ) / 2 = 23 ° . Which numbers below are closest to the magnitudes of the kinetic friction force and the component of the gravitational force parallel to the idealized inclined in-run? a. 30 N, 540 N b. 27 N, 540 N c. 12 N, 540 N d. 30 N, 230 N e. 27 N, 230 N f. 12 N, 230 N
Ski jumping in Vancouver The 2010 Olympic ski jumping competition was held at Whistler Mountain near Vancouver During a jump, a skier starts near the top of the in-run, the part down which the skier glides at increasing speed before the jump. The Whistler in-run is 116 m long and for the first part is tilted down at about 35 ° below the horizontal There is then a curve that transitions into a takeoff ramp, which is tilted 11 ° below the horizontal. The skier flies off this ramp at high speed body tilted forward and skis separated ( Figure 4.15 ). This position exposes a large surface area to the air, which creates lift, extends the time of the jump, and allows the jumper to travel farther In addition, the skier pushes off the exit ramp of the in-run to get a vertical component of velocity when leaving the ramp. The skier lands 125 m or more from the end of the in-run. The landing surface has a complex shape and is tilted down at about 35 ° below the horizontal. The skier moves surprisingly close (2 to 6 m) above the snowy surface for most of the jump. The coefficient of kinetic friction between the skis and the snow on the in-run is about 0.05 ± 0.02 , and skiers’ masses are normally small—about 60 kg. We can make some rough estimates about an idealized ski jump with an average in-run inclination of ( 35 ° + 11 ° ) / 2 = 23 ° . Which numbers below are closest to the magnitudes of the kinetic friction force and the component of the gravitational force parallel to the idealized inclined in-run? a. 30 N, 540 N b. 27 N, 540 N c. 12 N, 540 N d. 30 N, 230 N e. 27 N, 230 N f. 12 N, 230 N
Ski jumping in Vancouver The 2010 Olympic ski jumping competition was held at Whistler Mountain near Vancouver During a jump, a skier starts near the top of the in-run, the part down which the skier glides at increasing speed before the jump. The Whistler in-run is 116 m long and for the first part is tilted down at about
35
°
below the horizontal There is then a curve that transitions into a takeoff ramp, which is tilted
11
°
below the horizontal. The skier flies off this ramp at high speed body tilted forward and skis separated (Figure 4.15). This position exposes a large surface area to the air, which creates lift, extends the time of the jump, and allows the jumper to travel farther In addition, the skier pushes off the exit ramp of the in-run to get a vertical component of velocity when leaving the ramp. The skier lands 125 m or more from the end of the in-run. The landing surface has a complex shape and is tilted down at about
35
°
below the horizontal. The skier moves surprisingly close (2 to 6 m) above the snowy surface for most of the jump. The coefficient of kinetic friction between the skis and the snow on the in-run is about
0.05
±
0.02
, and skiers’ masses are normally small—about 60 kg. We can make some rough estimates about an idealized ski jump with an average in-run inclination of
(
35
°
+
11
°
)
/
2
=
23
°
.
Which numbers below are closest to the magnitudes of the kinetic friction force and the component of the gravitational force parallel to the idealized inclined in-run?
43. A mass må undergoes circular
motion of radius R on a hori-
zontal frictionless table, con-
nected by a massless string
through a hole in the table to
a second mass m² (Fig. 5.33).
If m₂ is stationary, find expres-
sions for (a) the string tension
and (b) the period of the circu-
lar motion.
m2
R
m₁
FIGURE 5.33 Problem 43
CH
70. A block is projected up an incline at angle 0. It returns to its initial
position with half its initial speed. Show that the coefficient of ki-
netic friction is μk = tano.
Passage Problems
A spiral is an ice-skating position in which the skater glides on one
foot with the other foot held above hip level. It's a required element
in women's singles figure-skating competition and is related to the
arabesque performed in ballet. Figure 5.40 shows Canadian skater
Kaetlyn Osmond executing a spiral during her medal-winning perfor-
mance at the 2018 Winter Olympics in Gangneung, South Korea.
77. From the photo, you can conclude
that the skater is
a. executing a turn to her left.
b. executing a turn to her right.
c. moving in a straight line out of
the page.
78. The net force on the skater
a. points to her left.
b. points to her right.
c. is zero.
79. If the skater were to execute the same
maneuver but at higher speed, the tilt
evident in the photo would be
a. less.
b. greater.
c. unchanged.
FIGURE 5.40 Passage
Problems 77-80
80. The tilt angle 0 that the skater's body
makes with the vertical is given ap-
proximately by 0 = tan¯¹(0.5). From this you can conclude…
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