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Practice 2A

source : k12.or.us

Practice 2A

Practice 2A

Problem Set 6.10: | 35
| 37 |
40 | 41 | 58 | 42
|
43 | 63 | 64 | 44
| 46 | 67 | 69 | 75
| Go up – by Cooper Spear and Chris Murray, 2002

35. Jane, looking at Tarzan, is running at top speed (5.6
m/s) and grabs a vine hanging vertically from a tall tree in the jungle. 
How high can she swing upward?  Does the length of the vine (or rope)
affect your answer? 

Tarzan wasn’t a ladies man he’d just come along and
scoop ’em up under his arm like that, quick as a cat in the jungle.
Jane, on the other hand, is a real lady, so let’s set this up.
This is a classic Conservation of energy problem:

Total Energy Before

=

Total Energy After

Before: Jane running along the level ground with a
velocity of 5.6 m/s. She has kinetic energy, but nothing else.

=

After: Jane having swung up to some height h, and
momentarily at rest. Here she has potential energy, and nothing else.

  1/2mv2

=

 mgDh

m cancels from both sides, giving   1/2mv2
=
mgDh
  1/2v2
=
gDh
(
1/2v2
)/g =
Dh1/2(5.6 m/s)2/(9.8
m/s/s) = 1.6 m
  
Assuming the vine is longer than 1.6 m, the length does not matter, the longer
the vine, the smaller the angle she swings through to reach a particular
height.(Table of contents)

37. A sled is initially given a shove up a frictionless
25.0 ° incline.  It reaches a maximum vertical height 1.35 m  higher
than where it started.  What was its initial speed? 

Total Energy Before

=

Total Energy After

Before: The sled moving along the ground before it
has gained any height.  It has kinetic energy and nothing else

=

After: The sled having slid up the plane and come to
a rest  It has potential energy and nothing else.

  1/2mv2

=

 mgDh

  1/2mv2

=

 m(9.8 m/s/s)(1.35 m)

m cancels from both sides giving:  1/2mv2
=
mgDh
  1/2v2
=
gDh
 v2
=
Ö{2gDh}
= {2(9.80 N/kg)(1.35 m)}
v =
5.14 m/s 
Thankfully, the degree of the incline is not relevant.(Table of contents)

40. A roller coaster, shown in Fig. 6-38, is pulled up to
point A where it and its screaming occupants are released from
rest.  Assuming no friction, calculate the speed at point B, C,
D.  C’s height is 25 m, D’s height is 12 m. 

In this problem the potential energy at point A
is turned into some combination of potential and kinetic at the other
points.  Arbitrarily, and conveniently, let’s decide that the potential
energy at Point B represents Zero Potential energy, as it is the lowest point.
Point B:

Total Energy Before

=

Total Energy After

Before: The rollercoaster is at point A, up at the top of the
hill at rest.  It has Potential energy, and nothing else.

=

After: The Rollercoaster is at the bottom of the
hill.  It has kinetic energy, but no potential, as we have
decided that the elevation at B is Zero.

 mgDh 

=

1/2mv2

Now, we don’t know the mass, but it is in
every term, so let’s just cancel it:mgDh = 1/2mv2
gDh = 1/2v2

2gDh = v2

v = Ö{2gDh}
= Ö{2(9.8
m/s/s)(30 m)} = 24.25 m/s = 24 m/s

Point C:

Total Energy Before

=

Total Energy After

Before: The rollercoaster is at point A, up at the top of the
hill at rest.  It has Potential energy, and nothing else.

=

After: The Rollercoaster is at Point C.  It has
both kinetic energy and potential energy.  (It is at height 25 m)

 mghA 

=

1/2mv2
+ mghC

Now, we don’t know the mass, but it is in
every term, so let’s just cancel it: mghA  =
1/2mv2
+ mghC
ghA  =
1/2v2
+ ghCghA – ghC 
=
1/2v2 

2(ghA – ghC) =  v2 
Ö{2g(hA –
hC)} =  vÖ{2(9.80
m/s/s)(30 m – 25 m)} =  v
v =
9.9 m/s

Point D

Total Energy Before

=

Total Energy After

Before: The rollercoaster is at point A, up at the top of the
hill at rest.  It has Potential energy, and nothing else.

=

After: The Rollercoaster is at Point D.  It has
both kinetic energy and potential energy.  (It is at height 12 m)

 mghA 

=

1/2mv2
+ mghD

Now, we don’t know the mass, but it is in
every term, so let’s just cancel it: mghA  =
1/2mv2
+ mghD
ghA  =
1/2v2
+ ghDghA – ghD  =
1/2v2 

2(ghA – ghD) =  v2 
Ö{2g(hA –
hD)} =  vÖ{2(9.80
m/s/s)(30 m – 12 m)} =  v
v =18.8 m/s =  19 m/s(Table of contents)

41. A projectile is fired at an upward angle of 45.0 °
from the top of a 265-m cliff with a speed of 185 m/s.   What will be
its speed when it strikes the ground below?  (Use conservation of energy)

Since we just want to know the speed, (and not the angle) we
don’t need to solve the projectile motion stuff we can look at it upon its
launch, and at the moment of impact with the ground.

Total Energy Before

=

Total Energy After

Before: The projectile is at the top of the 265 m
tall cliff, and moving at 185 m/s.  It has both potential and
kinetic.

=

After: The projectile is just striking the
ground.  It has no potential, only kinetic

1/2mv12 
+ mgh 

=

1/2mv22 

Now, we don’t know the mass, but it is in
every term, so let’s just cancel it: 1/2mv12  + mgh 
=
1/2mv22 
1/2v12  + gh 
=
1/2v22 
v12  + 2gh  =
v22 
Ö{v12 
+ 2gh} =  v2Ö{(185
m/s)2  + 2(9.8 m/s/s)(265 m)}
=  198.54 = 199 m/s(Table of contents)

58. How long will it take a 1750-W motor to lift a 285-kg
piano to a sixth-story window 16.0 m above? 

Assuming the lifting is done at a constant velocity. 
(i.e. the change in potential energy is the entire amount of energy
transformed) then we can use the formula for Power:
P = W/t – Where W is not just work as in W = Fs, but any transformation of
energy, so
W in this case = DEp
= mgDh = (285 J)(9.8 N/kg)(16.0 m) = 44688
J
Now if P = W/t, then t = W/P = (44688 J)/(1750 W) = 25.536 s = 25.5
s(Table of contents)

42. A 60-kg bungee jumper jumps from a bridge.  She
is tied to a 12-m-long bungee cord and falls a total of 31 m.  (a)
Calculate the spring constant k  of the bungee cord.  (b)
Calculate the maximum acceleration experienced by the jumper. 

When they reach the end of the Bungee cord, and begin to
stretch it, they have fallen 12 meters, and then they stretch it an additional
distance of 19 m.  (31 – 12 = 19) When they come to a halt, we know that
the total change in potential energy  (Due to the change in height equal
to 31 m) has been stored in the spring.

Total Energy Before

=

Total Energy After

Before: The jumper is on the bridge and has
potential energy and nothing else

=

After: The Bungee jumper has come to rest
stretching the cord distance s, and having fallen a total distance (s
+ 12m)  They have elastic potential and nothing else.

mgh 

=

1/2ks2 

So let’s solve for k:mgh  = 1/2ks2 
2mgh/s2 
= k = 2(60 kg)(9.80 m/s/s)(31 m)/(19 m)2 =
100.99 N/m = 1.0×102 N/m

Now we had better solve for the maximum upward acceleration of the
jumper.  It occurs at the maximum stretch of the bungee cord, as that is
when the stretch distance is the greatest.  The formula for the force
exerted by a spring is:
F = -kx = -(100.99 N/m)(-19 m) = 1918.74 N upwards
Now our expression of Newton’s second law has the weight downward (-) of (60
kg)(9.80 N/kg) = 588 N, and the force exerted by the bungee cord of 1918.74 N
upwards (+):
<1918.74 N – 588 N> = (60 kg)aa = 22.2 m/s/s = 22 m/s/s upwards(Table of contents)

43. A vertical spring (ignore its mass), whose spring
constant is 900 N/m, is attached to a table and is compressed 0.150 m.  (a)
What speed can it give to a 0.300-kg ball when released?  (b) How
high above its original position (spring compressed) will the ball fly? 

This has two parts, a) is from the lowest position, to the
ball just leaving the spring.  (which occurs .150 m above the initial
position)

Total Energy Before

=

Total Energy After

Before: The ball is on the spring, at rest and at
the lowest position.  It has elastic potential energy, but
nothing else.

=

After: The ball is just leaving the spring which
is fully extended.  It has potential and kinetic energy and
nothing else.

1/2ks2 

=

mgh  + 1/2mv2 

So let’s solve for velocity:1/2ks2  = mgh 
+ 1/2mv2 
1/2(900 N/m)(.150 m)2  = (.300
kg)(9.80 m/s/s)(.150 m)  + 1/2(.300 kg)v2 
v = 8.035 m/s = 8.03 m/s

Part b) is from the lowest position to the ball up in the air, having risen
some height h into the air.

Total Energy Before

=

Total Energy After

Before: The ball is on the spring, at rest and at
the lowest position.  It has elastic potential energy, but
nothing else.

=

After: The ball at its highest point, having risen
some distance “h” into the air.  It has gravitational potential
and nothing else.

1/2ks2 

=

mgh

So let’s solve for height:1/2ks2  = mgh
(1/2ks2)/mg = h{1/2(900 N/m)(.150 m)2}/(.300
kg)(9.80 m/s/s) = 3.44 m(Table of contents)

63. How much work can a 3.0-hp motor do in 1.0 h? 

3.0 hp = (3.0 hp)(745.7 W/hp) = 2237.1 W
1.0 h = 3600 s
P = W/t, so W = Pt = (2237.1 W)(3600 s) = 8053560 = 8.1 x 106
J(Table of contents)

64. A shot-putter accelerates a 7.3-kg shot from rest to
14 m/s.  If this motion takes 2.0 s, what average power was developed? 

The transformation of energy in this case is a change in
kinetic energy:
P = W/t = (1/2mv2 )/t
= (1/2(7.3 kg)(14 m/s)2 )/(2.0
s) = 357.7 W =
360 W(Table of contents)

44. A small mass m slides without friction along
the looped apparatus shown in Fig. 6-39.  If the object is to remain on the
track, even at the top of the circle (whose radius is r), from what
minimum height h must it be released? 

We know that at the top of the loop, the velocity must be
such that the centripetal acceleration is equal to “g”, so 
g = v2/r, so v2  = gr.  Since the formula for
kinetic energy uses v2, I will not take the square root.
We also know that at the top of the ramp, it has potential energy, and at the
top of the loop, it has both potential and kinetic:

Total Energy Before

=

Total Energy After

Before: The mass is at the top of the ramp at
rest.  It has gravitational potential energy, but nothing else.

=

After: The mass is at the top of the loop. 
It has both kinetic and potential energy, but nothing else.

mgh 

=

1/2mv2
+ mgh2

At the top of the loop, it is at a height of 2r, where r is
the radius of the loop: mgh  = 1/2mv2
+ mgh2 mgh  = 1/2mv2
+ mg2r
now v2  = gr, so mgh  = 1/2mgr + mg2r
Divide by mg:h  = 1/2r + 2r = 2.5r(Table of contents)

46. An engineer is designing a spring to be placed at
the bottom of an elevator shaft.  If the elevator cable should happen to
break at a height of h above the top of the spring, calculate the value
that the spring constant k should have so that passengers undergo an
acceleration of no more than 5.0 g when brought to rest.  Let M
be the total mass of the elevator and passengers. 

If you want the maximum upward acceleration to be 5.0 “g”s, then the
maximum upward force the spring exerts on the elevator would need to be 6
times the weight of the elevator.  (As 1 times the weight of the elevator
upward results in no upward acceleration).  If you prefer to be non
voodoo, let’s use Newton’s second law:  We have upward F the force of the
spring, and Mg the weight of the elevator.  so:
<F – Mg> = M(5g), and F = 6Mg.

Now if this upward force is from a spring compressed some distance s, the
maximum upward acceleration comes just as the spring is compressed to its
maximum distance.  then F = ks = 6Mg, so s = 6Mg/k

But what is the compression of the spring?  well, using energy, the
elevator starts at the top some distance h from the spring, and undergoes a
total change in height equal to h + s, so we have gravitational potential
energy turning into elastic potential of the spring:
Mg(h+s) =
1/2ks2 

And plugging in our expression for s:
Mg(h+6Mg/k) = 1/2k(6Mg/k)2 
Mgh + 6M2g2/k =  1/2k(36M2g2/k2)
Mgh + 6M2g2/k =  1/2k(36M2g2/k2)
gh + 6Mg2/k =  18Mg2/kgh = 12Mg2/kh = 12Mg/kk = 12Mg/h(Table of contents)

67. How fast must a cyclist climb a 6.0 ° hill to
maintain a power output of 0.25 hp?  Neglect work done by friction and
assume the mass of cyclist plus bicycle is 70 kg.

P = (.25 hp)(745.7 W/hp) = 186.425 W
P = W/t = Fs/t, since W = Fs
Now, s/t = v, so P = Fs/t = Fv, where v is the velocity.As the bicyclist climbs the hill, if we neglect friction, which is a
ridiculous notion, the only force opposing them is the component of gravity
parallel to the hill = mgsin(q)
So we have
P = Fv = mgsin(q)v,
and
v = P/mgsin(q)
= ( 186.425 W)/{(70 kg)(9.8 N/kg)sin(6.0o)} =
2.6 m/s(Table of contents)

69. Squaw Valley ski area in California claims that its
lifts can move 47,000 people per hour.  If the average lift carries people
about 200 m (vertically) higher, estimate the maximum total power needed.

First off, they need to change the name of the ski
area.  Squaw is not a polite reference to a woman.  It is a slang
reference to female genitalia.  (a good translation into English slang
starts with “c” and rhymes with “runt”)  I found this
out in a most embarrassing way when I was a teacher at an Indian school.

The power needed to just lift the people, if we assume that they have a mass
of about 70 kg (154 lbs on Earth)
Then P = W/t, and in this case, W = mgh, so
P = W/t = mgh/t = (47,000 people)(70 kg/person)(9.8 N/kg)(200 m)/(3600 s) =
1791222.222 W =
2 MW(Table of contents)

75. A .20 kg pinecone falls from a branch 18 m above the
ground.  (a) With what speed would it hit the ground if air resistance
could be neglected? (b) If it actually hits the ground with a speed of 10.0 m/s,
what was the average force of air friction exerted on it?

(a) This is simple, potential (mgh) turns all to kinetic (1/2mv2)

mgh =
1/2mv2
v =
Ö(2gh)
=
Ö(2gh)
=18.78 m/s = 19 m/s

(b) This is a little more subtle.

At an elevation of 18 m, the pinecone has mgh = (.20 kg)(9.8 N/kg)(18 m) =
35.28 J of energy

But, it arrives at the ground having only 1/2mv2
= 1/2(.20 kg)(10.0 m/s)2 
= 10. J of kinetic energy, which  means it lost 35.28 J – 10. J = 25.28 J
of energy to friction on the way down due to work against friction:

W = Fs
25.28 J = F(18 m), F = 1.4 N.  The book makes it negative as I
suppose.  Hmm, why would they do that.  Because the work was
negative?  (i.e. the air took energy away from the pinecone?)  It
would have been an upward force on the pinecone anyway…CJM

(Table of contents)

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