Page 39 - 2023-bfw-physics-stewart-3e-new.indd

7-6 Potential energy is energy related to reversible changes in a system’s configuration 321

∆K = K f − K i
1 1
= mv 2 − 0 = mv 2
2 f 2 f
(positive, because the skier is moving faster at the end of the ski
jump than at the beginning) So
1
+mg (H − D ) = mv 2
2 f
v 2 f = 2(gH − D )
v f = 2(gH − D )

Reflect

The answer does not involve the angles θ or φ, At starting point: v = 0
i
or any other aspect of the ramp’s shape. All that v i = 0
matters is the difference between the skier’s initial U grav,i = mgy i = 0
and final heights. If the ramp had a different At the point where skier y = H
i
v would be exactly
i
shape, the final speed f leaves the ramp: y = H v f
the same, as long as the difference in heights v f
− ),so
remained the same. U grav,f = mgy f = mg (HD y = D
f
y f = D
As we mentioned in the Watch Out! feature ∆U grav = U grav,f − U grav,i
just before this example, your answer also shouldn’t =−mg (H − D ) − 0 v i = 0
depend on our having chosen =y 0 to be at the v i = 0
low point of the ramp. For example, instead choose =−mg (H − D )
y = 0 to be where the skier leaves the ramp, and This is the same as with our
v should be the same. If =y
i
y i = H
the result for f 0 is the previous choice of = 0,y so y = H v v f
skier’s starting point, the end of the ramp is at we’ll find the same value f
y =−(H − D ) and you get the same ∆U grav . of .v f y = D
f
y = D
f
y = H – D
i
y = 0
f
NOW WORK Problems 2 and 5 in The Takeaway 7-6.

Spring Potential Energy

Our definition of potential energy is that a change in such energy must depend only
on the system’s final and initial configurations. For instance, the gravitational poten-
tial energy cannot depend on the path taken between the initial and final positions
of the object above Earth. Using gravitational potential energy, we saw how to quan-
tify the potential energy by comparing the change in potential energy when Earth was
in the system to the change in kinetic energy of an object due to work done by Earth
when it is outside of the system. We also saw that the amount of work that we would
have to do to lift an object between those two points was ∆U grav , the change in the
gravitational potential energy values for the Earth–object system as the object moved
between the two points. The same is true for the work done by an ideal spring. We
saw in Section 7-5 that the work you must do to stretch a spring from an extension
2
1
i x to an extension x is kx 2 f − kx (this is Equation 7-13, with 1 replaced by i and
1
f
2
2
i
2 replaced by f). Note that this also depends on only the initial and final extensions of
the spring, not on the details of how you got the end of the spring from one position
to the other. In this case, the spring is our system. If you fix one end of the spring, and
pull on the other end, it stretches and if you hold it at a given stretch it has no kinetic
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.

08_stewart3e_33228_ch07_284_333_8pp.indd 321 20/08/22 8:46 AM