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320 Chapter 7 Conservation of Energy and an Introduction to Energy and Work

EXAMPLE 7-8 A Ski Jump

A skier of mass m starts at rest at the top of a ski jump ramp ( Figure 7-19 ). The
,
vertical distance from the top of the ramp to the lowest point is H and the vertical Skier ,
Skier, mass m
Skier
.
distance from the lowest point to where the skier leaves the ramp is D The first part v = 0
i

of the ramp is at an angle θ from the horizontal, and the second part is at an angle φ.
Derive an expression for the speed of the skier when she leaves the ramp in terms of H v f = ?
=
f f
the variables given in the problem and appropriate physical constants. Assume that q
her skis are well waxed, so that there is negligible friction between the skis and the f f f f D
ramp, and ignore air resistance.
Figure 7-19 Flying off the ramp
What is the skier’s speed when she
leaves the ramp, neglecting friction?
Set Up
If we neglect friction, the only forces that are Work-energy theorem: F n
exerted on the skier are the gravitational force
K
and the normal force exerted by the ramp. The W =∆ +∆U grav +∆E internal (7-12)
normal force does no work on the skier because Gravitational potential energy:
it is always exerted perpendicular to the ramp

and so is perpendicular to her direction of U = mgy (7-14)
grav
motion. We are not worried about the skier’s Kinetic energy:
own internal energy to keep herself warm, and so
on; just the energy that affects her motion, and = 1 mv (7-8) mg
2
K
she just slides down the slope, so we can neglect 2
her internal energy. The gravitational force does
no work if we choose our system to be the Earth–
,
skier system. So we can use Equation 7-12 with
W = 0 and ∆E internal = 0 to calculate the change

in the skier’s kinetic energy and hence her final
v .
speed f
Solve
Let’s take = 0 to be at the low point of the ramp. At starting point:
y

K i = 0

y
v
i
The skier then begins at rest ( i = 0) at = H 1 U = mgH
i
and is at =y D when she leaves the ramp. Use i K = mv 2 = 0

f
Equations 7–8, 7-12, and 7-14 to solve for .v 2 i 2
f
i
U grav,i = mgy i = mgH y = H K f f = ½ mv f
U = mgD
At the point where the skier
f
leaves the ramp: y = D
1
K = mv 2
f 2 f
U grav,f = mgy f = mgD
Use Equation 7-12 :
W =∆ +∆U grav +∆E internal
K
W = 0and ∆E internal = 0
=∆ +∆K
0 U grav
,

−∆U grav = ∆K here
∆U grav = U grav,f − U grav,i = mgD − mgH
= mg (D − H ) = −mg (H − D )
(negative, because <D H ) and

Uncorrected proofs have been used in this sample. Copyright © Bedford, Freeman & Worth Publishers.
Distributed by Bedford, Freeman & Worth Publishers. For review purposes only. Not for redistribution.
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