Transcript 6-7 Problem Solving Using Conservation of Mechanical Energy For

6-7 Problem Solving Using Conservation of
Mechanical Energy
In the image on the left, the total
mechanical energy is:
The energy buckets (right)
show how the energy
moves from all potential to
all kinetic.
Example 6-8
If the original height of the rock is y1=h=3.0 m, calculate the rock’s speed
when it has fallen to 1.0 m abve the ground.
y1 = 3.0 m, v1 = 0, y 2 = 1.0 m, v 2 = ?
1 2
1
mv 1 + mgy 1 = mv 22 + mgy 2
2
2
v 22 = 2g(y 1 - y 2 ) = 2(9.8 m/s 2 )(3.0 m -1.0 m) = 39.2 m2 /s 2
v 2 = 39.2 m2 /s 2 = 6.3 m/s

6-7 Problem Solving Using Conservation of
Mechanical Energy
If there is no friction, the speed of a roller
coaster will depend only on its height
compared to its starting height.
Example 6-9
Assuming the height of the hill is 40 m, and the roller-coaster ca starts from
rest at the top, calculate (a) the speed of the roller coaster car at the bottom
of the hill, and (b) at what height it will have half this speed. Take y=0 at the
bottom of the hill.
(a) y1 = 40 m, v1 = 0, y 2 = 0
1 2
1
mv 1 +mgy 1 = mv 22 +mgy 2
2
2
1
mgy 1 = mv 22
2
v 2 = 2gy1 = 2(9.80 m/s 2 )(40 m) = 28 m/s
(b) y1 = 40 m, v1 
= 0, y 2 = ?, v 2 = 14 m/s
1 2
1
mv 1 + mgy 1 = mv 22 + mgy 2
2
2
(14 m/s) 2
v 22
y 2 = y1 = 40 m = 30 m
2g
2(9.80 m/s 2 )
6-7 Problem Solving Using Conservation of
Mechanical Energy
For an elastic force, conservation of energy tells
us:
(6-14)
Example 6-11
A dart of mass 0.100 kg is pressed against the spring of a toy dart gun.
The spring (with spring stiffness constant k=250 N/m) is compressed 6.0
cm and released. If the dart detaches from the spring when the spring
reaches its natural length (x=0), what speed does the dart acquire?
x1 = -0.060 m, v1 = 0, x 2 = 0, v 2 = ?
1
1
KE1 = 0, PE1 = kx12 , KE 2 = mv 22 , PE 2 = 0
2
2
1
1
0 + kx12 = mv 22 + 0
2
2
kx12 (250 N/m)(-0.060 m) 2
2
v2 =
=
= 9.0 m2 /s 2
m
(0.100 kg)
v 2 = v 22 = 9.0 m2 /s 2 = 3.0 m/s

Example 6-12
A ball of mass m=2.60 kg, starting from rest, falls a vertical distance h=55.0
cm before striking a vertical coiled spring, which it compresses an amount
Y=15.0 cm. Determine the spring stiffness constant of the spring. Assume
the spring has negligible mass, and ignore air resistance. Measure all
distances from the point where the ball first touches the uncompressed
spring (y=0 at this point).
Part 1 : ball falls height h : y1 = h = 0.550 m, y 2 = 0
1 2
1
1
mv 1 +mgy 1 = mv 22 +mgy 2  0 +mgh = mv 22 + 0
2
2
2
v 2 = 2gh = 2(9.80 m/s 2 )(0.550 m) = 3.28 m/s
Part 2 : ball compresses spring : E(ball touches spring) = E(spring compresses)
1 2
1
1
1
mv 2 +mgy 2 + ky 22 = mv 23 +mgy 3 + ky 23
2
2
2
2
y 2 = 0, v 2 = 3.28 m/s, v 3 = 0,y = -Y = -0.150 m
1 2
1
mv 2 +0 +0 = 0 - mgY + kY2
2
2

1

2 1
2
2
2
k = 2  mv 22 +mgY =
(2.60
kg)(3.28
m/s)
+(2.60
kg)(9.80
m/s
)(0.150
m)
=1590 N/m

 (0.150 m) 2 
2

Y 2
6-8 Other Forms of Energy; Energy
Transformations and the
Conservation of Energy
Some other forms of energy:
Electric energy, nuclear energy, thermal energy,
chemical energy.
Work is done when energy is transferred from
one object to another.
Accounting for all forms of energy, we find that
the total energy neither increases nor
decreases. Energy as a whole is conserved.
6-9 Energy Conservation with Dissipative
Processes; Solving Problems
If there is a nonconservative force such as
friction, where do the kinetic and potential
energies go?
They become heat; the actual temperature rise of
the materials involved can be calculated.
6-9 Energy Conservation with Dissipative
Processes; Solving Problems
Problem Solving:
1. Draw a picture.
2. Determine the system for which energy will
be conserved.
3. Figure out what you are looking for, and
decide on the initial and final positions.
4. Choose a logical reference frame.
5. Apply conservation of energy.
6. Solve.
Example 6-13
The roller-coaster car reaches a vertical height of only 25 m on the second
hill before coming to a momentary stop. It traveled a total distance of 400 m.
Estimate the average friction force (assume constant) on the car, whose
mass is 1000 kg.
v1 = 0, y1 = 40 m, v 2 = 0, y 2 = 25 m, d = 400 m
1 2
1
mv 1 +mgy 1 = mv 22 +mgy 2 +Ffr d
2
2
0 +(1000 kg)(9.80 m/s 2 )(40 m) = 0 +(1000 kg)(9.80 m/s 2 )(25 m) +Ffr (400 m)
Ffr = 370 N
6-10 Power
Power is the rate at which work is done –
(6-17)
In the SI system, the units of
power are watts:
The difference between walking
and running up these stairs is
power – the change in
gravitational potential energy is
the same.
Example 6-14
A 60. Kg jogger runs up a long flight of stairs in 4.0 s. The vertical
height of the stairs is 4.5 m. (a) Estimate the jogger’s power output in
watts and horsepower. (b) How much energy did this require?
W mgy (60 kg)(9.80 m/s 2 )(4.5 m)
(a) P =
=
=
= 660 W
t
t
4.0 s
There 746 W in 1 hp, so the jogger' s doing work at a
rate just under 1 hp, which humans can' t keep up for long.
(b) E = Pt = (660 W)(4.0 s) = 2600 J
The person had to transform more energy than this 2600 J. The total
 energy transformed by a person or an engine always includes some
thermal energy (recall how hot you get running up stairs).
6-10 Power
Power is also needed for acceleration and for
moving against the force of gravity.
The average power can be written in terms of the
force and the average velocity:
(6-17)
Example 6-15
Calculate the power required of a 1400 kg car under the following
circumstances: (a) the car climbs a 10 degree hill at a steady 80.
Km/h; and (b) the car accelerates along a level road from 90. to 110
km/h in 6.0 s to pass another car. Assume the retarding force on the
car is FR=700 N (due to air resistance).
(a) F = 700 N +mgsin10
= 700 N +(1400 kg)(9.80 m/s 2 )(0.174) = 3100 N
v = 80. km/h = 22 m/s
P = Fv = (3100 N)(22 m/s) = 6.80x10 4 W = 91 hp
 30.6 m/s - 25.0 m/s
(b) ax =
= 0.93 m/s 2
6.0 s
ma x = Fx = F - FR
F = ma x +FR = (1400 kg)(0.93 m/s 2 ) + 700 N = 2000 N
P = Fv = (2000 N)(30.6 m/s) = 6.12x10 4 W = 82 hp
Summary of Chapter 6
• Work:
•Kinetic energy is energy of motion:
• Potential energy is energy associated with forces
that depend on the position or configuration of
objects.
•
•The net work done on an object equals the change
in its kinetic energy.
• If only conservative forces are acting, mechanical
energy is conserved.
• Power is the rate at which work is done.
Homework - Ch. 6
• Questions #’s 2, 3, 4, 12, 13, 14, 21, 24
• Problems #’s 5, 9, 19, 29, 31, 37, 39,
43, 49, 63, 65, 67, 69