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RECITATION
WEEK 10
CHAPTER 7
7.27. A spring of force constant 300.0 N/m and unstretched length 0.240 m is
stretched
by two forces, pulling in opposite directions at opposite ends of the spring,
that increase
to 15.0 N. How long will the spring now be, and how much work was required to
stretch
it that distance?
Set Up: Solve Fon spring  kx for x to determine the length of stretch and use
Won spring   12 kx 2 to assess the corresponding work.
Solve: x 
Fon spring
k

15.0 N
 0.0500 m
300.0 N/m
The new length will be 0.240 m  0.0500 m  0.290 m. The corresponding work done
is,
Won spring 


1
300.0 N/m 0.0500 m
2
  0.375 J
2
Reflect: Fon spring is always the force applied to one end of the spring, thus
we did not
need to double the 15.0 N force. Consider a free-body diagram of a spring at
rest; forces
of equal magnitude and opposite direction are always applied to both ends of
every
section of the spring examined.
7.31. How high can we jump? The maximum height a typical human can jump from a
crouched start is about 60 cm. By how much does the gravitational potential
energy
increase for a 72 kg person in such a jump? Where does this energy come from?


Solve: U grav  mg( y f  yi )  72kg 9.80m / s 2 0.60m  420 J
Reflect: This gravitational potential energy comes from elastic potential stored
in his
tensed muscles.
7.36. A certain spring stores 10.0 J of potential energy when it is stretched by
2.00 cm
from its equilibrium position. (a) How much potential energy would the spring
store if it
were stretched an additional 2.00 cm? (b) How much potential energy would it
store if it
were compressed by 2.00 cm from its equilibrium position? (c) How far from the
equilibrium position would you have to stretch the spring to store 20.0 J of
potential
energy? (d) What is the force constant of this spring?
Set Up: Use U el  12 kx 2 for the potential energy. U 1 for x1 and U 2 for x2
are related by


2
U 2 / U1  x2 / x1 .
x
Solve: (a) U 2  U 1  2
 x1
2
2

 4.00  10  2 m 
  10.0 J 
  10.0 J 4.00 J   40.0 J
2
2
.
00

10
m



(b) U el depends only on the magnitude of x, not on its sign; therefore, the
elastic potential
energy is U el  10.0 J.
(c) x2  x1
(d) k 




U2
20.0 J
 2.00  10 2 m
 2.00 2.00  10 2 m 100cm / m  2.83cm
U1
10.0 J
2U
210.0 J 

2
x
2.00  10 2 m


2
 5.00  10 4 N / m
7.45. Pebbles of weight w are launched from the edge of a vertical cliff of
height h at
speed v0. How fast (in terms of the quantities just given) will these pebbles be
moving
when they reach the ground if they are launched (a) straight up, (b) straight
down, (c)
horizontally away from the cliff, and (d) at an angle θ above the horizontal?
(e) How
would the answers to previous parts change if the pebbles weighed twice as much?
1 w  2
1
2
 g v f and K i   w g v0 . Let




2
2
1 w  2
 
v0  wh .
2 g
Set Up:. Use K f  U f  K i  U i with K f 
yi  h and yf  0; this gives
1 w  2

v f
2 g
Solve: (a)-(d) v f  v0  2 gh (e) Since K and U are both proportional to the
weight
2
w  mg, w divides out of the expression and v f is unaffected by a change in
weight.
Reflect: The initial kinetic energy depends only on the initial speed and is
independent of
the direction of the initial velocity.
7.49. Two stones of different masses are thrown straight upward, one on earth
and one
on the moon (where gravity produces an acceleration of 1.67 m/s2). Both reach
the same
height. If the stone on the moon had an initial speed v, what was the initial
speed of the
stone on earth, in terms of v?
Set Up: Apply K f  U f  K i  U i to the motion of the stone on the moon with
U i  0
and K f  0 at the maximum height. The conservation equation becomes U f  K i
with
1
vi  v moon,i  v and K i  K moon,i  mv 2 . Setting yf  h and yi  0 gives
2
2
1 2
v
mv  mgh . Note that
is the same on the earth and moon and let subscript 1 refer
2
g
to the moon and subscript 2 refer to the earth.
Solve: v1 / g1  v 2 / g 2 , v 2  v1
2
2
g2
9.80m / s 2
 v1
 2.42v1
g1
1.67m / s 2
Reflect: Gravity is stronger on earth so a larger initial speed is needed in
order to achieve
the same height as on the moon.
7.60. A 1.1 kg mass hangs from a spring of force constant 400.0 N/m. The mass is
then
pulled down 13.0 cm from equilibrium position and released. At the end of five
complete
cycles of vibration, the mass reaches only 10.0 cm from the equilibrium
position. (a)
How much mechanical energy is lost during these five cycles? (b) What percentage
of
mechanical energy is lost during the five cycles?
Set Up: Applying Emech  K  U with K i  K f  0, the change in mechanical
energy is
simply the difference between the initial and final elastic potential energy of
the stretched
spring: Emech,i  Emech,f  12 k xi2  xf2 . Also define the initial position as
the point of


release and the final position as the lowest position after five complete
cycles.
Solve: (a) Emech,i  Emech,f 
(b) The percentage lost is
E
1
2
400.0 N/m 0.130 m   0.100 m    1.38 J.
2
mech ,i
 E mech, f
E mech,i
2
 1.38 J 

 100%  40.8%
3.38 J 
Reflect: Note that the spring, in its equilibrium position, is stretched by the
weight of the
hanging mass. Consequently, an alternative solution can be developed that
explicitly
incorporates gravitational potential energy. Let d be the distance the spring is
stretched at
equilibrium due to the weight of the object.
F  kd  mg
or



1.1 kg 9.80 m/s 2
mg
d

 2.70 3 102 m  0.027 m
k
400 N/m
Define a new reference point; let y  yi  0 where the mass is 0.130 m below the
equilibrium point. At this point, the spring is stretched
di  0.130 m  0.027 m  0.157 m. The total potential energy is therefore:
Emech,i  U el  U grav  12 kdi 2  mgyi 
1
2
400 N/m 0.157 m   0  4.93 J
2
Then when the mass is 0.100 m below the equilibrium position, y  yf  0.030 m
and
the spring is stretched df  0.100 m  0.027 m  0.127 m. The total energy is
then:
Emech,f  U el  U grav  12 kdf 2  mgyf 
1
2
400 N/m 0.127 m   1.1 kg 9.80 m/s 0.030 m  3.55 J
2
2
The change in mechanical energy is thus 4.93 J  3.55 J  1.38 J, the same as
found in
the simplified calculation of part (a).
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