Unformatted text preview: edge of the cliff. In the second part she leaves the snow and falls freely toward
the ground. We can again employ the workenergy theorem to find her speed just before she
lands.
SOLUTION The drawing at the right shows the
three forces that act on the skier as she glides on the
snow. The forces are: her weight mg, the normal force
FN, and the kinetic frictional force fk. Her
displacement is labeled as s. The workenergy
theorem, Equation 6.3, is
2
W = 1 mvf2 − 1 mv0
2
2 FN +y fk
s
65.0°
25.0° mg where W is the work done by the net external force that acts on the skier. The work done by
each force is given by Equation 6.1, W = ( F cos θ ) s , so the workenergy theorem becomes 300 WORK AND ENERGY cos
( mg cos 65.0o ) s + ( f k4 180o ) s + ( FN cos 90o 4 = 1 mv 2 − 1 mv 2
)s
144444444424444444443 2 f 2 0
W Since cos 90° = 0, the third term on the left side can be eliminated. The magnitude fk of the
kinetic frictional force is given by Equation 4.8 as f k = µ k FN . The magnitude FN of the
normal force can be determined by noting that the skier does not leave the surface of the
slope, so ay = 0 m/s2. Thus, we have that ΣFy = 0, so
FN − mg cos 25.0° = 0
1444
2444
3 or FN = mg cos 25.0° ΣFy The magnitude of the kinetic frictional force becomes f k = µ k FN = µ k mg cos 25.0o .
Substituting this result into the workenergy theorem, we find that mg
s2
( mg cos 65.0o ) s + ( µ4 cos 25.0o ) ( cos 180o )3 = 1 mvf2 − 1 mv02
2
144444444k 2444444444
W Algebraically eliminating the mass m of the skier from every term, setting cos 180° = –1
and v0 = 0 m/s, and solving for the final speed vf gives
vf = 2 gs ( cos 65.0o − µ k cos 25.0o )
= 2 ( 9.80 m/s 2 ) (10.4 m ) cos 65.0o − ( 0.200 ) cos 25.0o = 7.01 m/s The drawing at the right shows her displacement s during free
fall. Note that the displacement is a vector that starts where she
leaves the slope and ends where she touches the ground. The
only force acting on her during the free fall is her weight mg.
The workenergy theorem, Equation 6.3, is 3.50 m θ s
mg 2
W = 1 mvf2 − 1 mv0
2
2 The work W is that done by her weight, so the workenergy theorem becomes
s
( mg cosθ )3 = 1 mvf2
14 244 2
4
W 2
− 1 mv0
2 In this expression θ is the angle between her weight (which points vertically downward)
and her displacement. Note from the drawing that s cos θ = 3.50 m. Algebraically
eliminating the mass m of the skier from every term in the equation above and solving for
the final speed vf gives Chapter 6 Problems 301 2
vf = v0 + 2 g ( s cos θ ) = ( 7.01 m/s ) 2 + 2 ( 9.80 m/s 2 ) ( 3.50 m ) = 10.9 m/s ______________________________________________________________________________
29. REASONING The change in gravitational potential energy for both the adult and the child
is ∆PE = mghf − mgh0, where we have used Equation 6.5. Therefore, ∆PE = mg(hf − h0). In
this expression hf − h0 is the vertical height of the second floor above the first floor, and its
value is not given. However, we know that it is the same for both staircases, a fact that will
play the central role in our solution.
SOLUTION Solving ∆PE = mg(hf − h0) for hf − h0, we obtain hf − h0 = ( ∆PE )Adult
mAdult g and hf − h0 = ( ∆PE )Child
mChild g Since hf − h0 is the same for the adult and the child, we have ( ∆PE )Adult ( ∆PE )Child
= mAdult g mChild g Solving this result for (∆PE)Child gives ( ∆PE )Child = ( ∆PE )Adult mChild ( 2.00 ×103 J ) (18.0 kg )
mAdult = 81.0 kg = 444 J 30. REASONING At a height h above the ground, the gravitational potential energy of each
clown is given by PE = mgh (Equation 6.5). At a height hJ, Juggle’s potential energy is,
therefore, PEJ = mJghJ, where mJ is Juggles’ mass. Similarly, Bangles’ potential energy can
be expressed in terms of his mass mB and height hB: PEB = mBghB. We will use the fact that
these two amounts of potential energy are equal (PEJ = PEB) to find Juggles’ mass mJ. SOLUTION Setting the gravitational potential energies of both clowns equal, we solve for
Juggles’ mass and obtain mJ g hJ = mB g hB
123 1 24
43
PE J PE B or mJ = mB hB
hJ = ( 86 kg ) ( 2.5 m ) =
( 3.3 m ) 65 kg 302 WORK AND ENERGY 31. SSM REASONING During each portion of the trip, the work done by the resistive force
is given by Equation 6.1, W = ( F cos θ )s . Since the resistive force always points opposite
to the displacement of the bicyclist, θ = 180° ; hence, on each part of the trip,
W = ( F cos 180 ° )s = − Fs . The work done by the resistive force during the round trip is the
algebraic sum of the work done during each portion of the trip.
SOLUTION
a. The work done during the round trip is, therefore,
Wtotal = W1 + W2 = − F1s1 − F2 s2 = − (3.0 N)(5.0 × 103 m) − (3.0 N)(5.0 × 103 m) = –3.0 × 10 4 J
b. Since the work done by the resistive force over the closed path is not zero, we can
conclude that the resistive force is not a conservative force .
______________________________________________________________________________
32. REASONING AND SOLUTION
a. From the definition of work, W = Fs cos θ. For upward motion θ = 180° so
W = – mgs =...
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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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