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 6-8If 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.

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y1 = 3.0 m, v1 = 0, y2 =1.0 m, v2 = ?12

mv12 +mgy1 = 1

2mv2

2 +mgy2

v22 = 2g(y1 - y2) = 2(9.8 m/s2)(3.0 m -1.0 m) = 39.2 m2/s2

v2 = 39.2 m2/s2 = 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 6-9Assuming 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.

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(a) y1 = 40 m, v1 = 0, y2 = 0

12

mv12 +mgy1 = 1

2mv2

2 +mgy2

mgy1 = 12

mv22

v2 = 2gy1 = 2(9.80 m/s2)(40 m) = 28 m/s

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(b) y1 = 40 m, v1 = 0, y2 = ?, v2 =14 m/s

12

mv12 +mgy1 = 1

2mv2

2 +mgy2

y2 = y1 - v22

2g= 40 m -

(14 m/s)2

2(9.80 m/s2)= 30 m

6-7 Problem Solving Using Conservation of Mechanical Energy

For an elastic force, conservation of energy tells us:

(6-14)

Example 6-11A 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?

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x1 = -0.060 m, v1 = 0, x2 = 0, v2 = ?

KE1 = 0, PE1 = 12

kx12, KE 2 = 1

2mv2

2, PE2 = 0

0 + 12

kx12 = 1

2mv2

2 +0

v22 = kx1

2

m= (250 N/m)(-0.060 m)2

(0.100 kg)= 9.0 m2/s2

v2 = v22 = 9.0 m2/s2 = 3.0 m/s

Example 6-12A 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).

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Part 1: ball falls height h : y1 = h = 0.550 m,y2 = 012

mv12 +mgy1 = 1

2mv2

2 +mgy2 → 0 +mgh = 12

mv22 +0

v2 = 2gh = 2(9.80 m/s2)(0.550 m) = 3.28 m/s

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Part 2 : ball compresses spring : E(ball touches spring) = E(spring compresses)12

mv22 +mgy2 + 1

2ky2

2 = 12

mv32 +mgy3 + 1

2ky3

2

y2 = 0, v2 = 3.28 m/s, v3 = 0,y = -Y = -0.150 m12

mv22 +0 +0 = 0 - mgY + 1

2kY2

k = 2Y2

12

mv22 +mgY

⎡ ⎣ ⎢

⎤ ⎦ ⎥= 2

(0.150 m)2

12

(2.60 kg)(3.28 m/s)2 +(2.60 kg)(9.80 m/s2)(0.150 m) ⎡ ⎣ ⎢

⎤ ⎦ ⎥=1590 N/m

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 6-13The 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.

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v1 = 0, y1 = 40 m, v2 = 0, y2 = 25 m, d = 400 m12

mv12 +mgy1 = 1

2mv2

2 +mgy2 +Ffrd

0 +(1000 kg)(9.80 m/s2)(40 m) = 0 +(1000 kg)(9.80 m/s2)(25 m) +Ffr (400 m)Ffr = 370 N

6-10 PowerPower is the rate at which work is done –

The difference between walking and running up these stairs is power – the change in gravitational potential energy is the same.

(6-17)

In the SI system, the units of power are watts:

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?

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(a) P = Wt

= mgyt

= (60 kg)(9.80 m/s2)(4.5 m)4.0 s

= 660 W

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 = P t = (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).

Example 6-14

6-10 PowerPower 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 6-15Calculate 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).

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(a) F = 700 N +mgsin10

= 700 N +(1400 kg)(9.80 m/s2)(0.174) = 3100 N v = 80. km/h = 22 m/s

P = Fv = (3100 N)(22 m/s) = 6.80x104 W = 91 hp

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(b) a x = 30.6 m/s - 25.0 m/s6.0 s

= 0.93 m/s2

max = ΣFx = F - FR

F = max +FR = (1400 kg)(0.93 m/s2) + 700 N = 2000 N

P = Fv = (2000 N)(30.6 m/s) = 6.12x104 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