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LEARNING UNIT 11 Work and Energy Chapter 6

Work and Energy

Study the indicated material in the textbook. 6.1 6.2 6.3 6.4 6.5 6.6 6.7 6.10

pages 148 151 155 157 159 164 165 169

Work Done by a Constant Force The Work- Energy Theorem and Kinetic Energy Gravitational Potential Energy Conservative versus Non-conservative Forces The Conservation of Mechanical Energy Non-conservative Forces and the Work-Energy Theorem Power Concept & Calculations

After completion of this learning unit you should be able to achieve the following outcomes: Demonstrate knowledge, comprehensive and acquired skills regarding energy and work.

Work W=Fs cosθ

F = Magnitude of the force (N) = Angle between the direction of the force and the direction of the displacement s = Magnitude of the displacement (m)

Calculate the work done in each of the following cases: Force = 10 N

Force = 10 N o

30 Displacement = 2 m

Displacement = 2 m

Force = 10 N Displacement = 2 m Force = 10 N Displacement = 2 m Force = 10 N Force = 10 N o

30

Displacement = 2 m

Displacement = 2 m 59 | P a g e

Kinetic energy: Ability to do work because an objects motion. 1 2 Ek Ek = Kinetic energy (J) 2m v m = mass (kg) v = speed (m.s-1) Potential energy Ability to do work because of an objects position in a gravitational field. Ep=mgh Ep = Potential energy m =mass g = gravitational acceleration h = height The work-energy theorem: When a net external force does work on an object, the kinetic energy of the object changes from its initial value of Eko to its final value of Ekf. The difference between the two values is equal to the work done

Wnet

Ek

1

2 mv

f

1

2

2 mvi

2

In the above equation, the work done by weight is included in the left side of the equation. Another form of the above equation is below when the work done by weight is the change in the potential energy.

Wnet Wc Wnc

Ek

1 mv 2

f

2

mgh

f

1 mv 2 i

2

mghi

Conservative force A force is conservative if the work it does on a moving object is independent of the path of motion of the object. Principle of conservation of mechanical energy: The total mechanical energy (E = Ek + Ep) of an object remains constant as the object moves, provided that the net work done by external nonconservative forces is 0 (W nc = 0) Principle of conservation of energy: Energy can neither be created nor destroyed, but can only be converted from one from to another.

E1

E2

1 mv 2 mgh 2 i

i

1 mv 2 f 2

mgh f

Power The rate of doing work.

P

W t

P = Power (W) W = Work (J) t = Time (s)

When you finish this chapter, you should be able to 1. Define (a) work, (b) joule, (c) power, (e) kinetic energy (f) work-energy theorem, (g) gravitational potential energy, (h) law of conservation of energy 2. Compute the work done on an object by a specified force when the object is moved through a given distance. 3. Compute power in simple situations. 4. Compute the change in gravitational potential energy of an object as it is moved from one place to another. 5. Distinguish between conservative and non-conservative forces 6. Use the work-energy theorem to calculate unknowns. 60 | P a g e

LEARNING UNIT 11 EXERCISE WITH REASONING AND SOLUTIONS: 1. A 1200-kg car is being driven up a 5.0° hill. The frictional force is directed opposite to the motion of the car and has a magnitude of

. A force

is applied to the car by the road and propels the car forward. In

addition to these two forces, two other forces act on the car: its weight and the normal force perpendicular to the road surface. The length of the road up the hill is 290 m. What should be the magnitude of , so that the net work done by all the forces acting on the car is +150 kJ?

directed

REASONING AND SOLUTION The net work done on the car is: WT = WF + Wf + Wg + WN WT = Fs cos 0.0° + f s cos 180° – mgs sin 5.0° + FNs cos 90° Rearranging this result gives

WT s

F

=

mg sin 5.0°

f

150 103 J

2 3 524 N + 1200 kg 9.80 m/s sin 5.0° 2.07 10 N 290 m

_______________________________________________________________________________________ 2. The skateboarder in the drawing starts down the left side of the ramp with an initial speed of 5.4 m/s. If nonconservative forces, such as kinetic friction and air resistance, are negligible, what would be the height h of the highest point reached by the skateboarder on the right side of the ramp?

REASONING The distance h in the drawing in the text is the difference between the skateboarder’s final and initial heights (measured, for example, with respect to the ground), or h = hf h0. The difference in the heights can be determined by using the conservation of mechanical energy. This conservation law is applicable because nonconservative forces are negligible, so the work done by them is zero (Wnc = 0 J). Thus, the skateboarder’s final total mechanical energy Ef is equal to his initial total mechanical energy E 0: 2 2 1 1 2

mv

mgh

f

1

f

1

442 4 43

2

0

1v2

h f0h

(6.9b)

0

442 4 43

E0

Ef Solving Equation 6.9b for h f h0 , we find that

mgh

mv

1 v2 2 f

2 0

g

142 43

h SOLUTION Using the fact that v0 = 5.4 m/s and vf = 0 m/s (since the skateboarder comes to a momentary rest), the distance h is 1

h

v2

1

2 0

2 f

g

2 v

1 5.4 2

m/s

10 2

2 9.80 m/s

m/s

2

1.5 m

__________________________________________________________________________________________ 61 | P a g e

3. A 3.00-kg model rocket is launched straight up. It reaches a maximum height of 1.00 x 10 2 m above where its engine cuts out, even though air resistance performs – 8.00 x 10 2 J of work on the rocket. What would have been this height if there were no air resistance? REASONING AND SOLUTION For the actual motion of the rocket KE = Wnc – PE = Wnc – mgh

2

(6.7b)

2

2

KE = –8.00 x 10 J – (3.00 kg)(9.80 m/s )(1.00 x 10 m) = –3740 J Since KE = KEf – KE0 = – KE0 , the initial kinetic energy of the rocket is then KE0 = 3740 J If the rocket were launched with this initial kinetic energy and no air resistance, the potential energy of the rocket at the top of its trajectory would be PE = mgh = 3740 J Hence,

3740 J h

127 m

2 3.00 kg 9.80 m/s

4. A top fuel dragster with a mass of 500.0 kg starts from rest and completes a quarter mile (402 m) race in a time of 5.0 s. The dragster's final speed is 130 m/s. Neglecting friction, what average power was needed to produce this final speed? REASONING AND SOLUTION 1 1 W12 mv f 2 2 500(130) 2 mvi 2

W t

P

1 2 (500)(0)

2

2

4225000 J

4225000 845000 W 5

5. In a truck-loading station at a post office, a 2 kg package is released from rest at point A on a frictionless track that is a quadrant of a circle of radius 1 m. The package slides from point A to point B on a frictionless surface. From point B it slides on a different horizontal surface (not frictionless) where it comes to rest at point C. Calculate: A 1m 2m B

a. the speed at point B. b. the friction coefficient between B and C.

C

REASONING AND SOLUTION (a)

E f E0 1 vf2 2 (b)

1 mv 2 mgh 1 mv 2 f f 0 2 2 1 2 9.8 0 (0) 9.8 1 v f 2

N

y

F

FW 0

4.43m / s

v2 u2 2as 2

Fmg N

FN

mgh0

9.8 2 19.6N

0 4.43 2 a 2 a 4.91m / S 2

F ma x

f k ma k FN k

ma ma

2 ( 4.91)

FN

19.6

0.50

62 | P a g e

Conceptual questions: 1, 6, 8 Problems: 1, 4, 7(also determine what was the total work done), 7, 15, 22, 34, 36, 37, 38, 40, 58, 61, 68 Tutorial questions: 6. A 10.0-g bullet traveling horizontally at 755 m/s strikes a stationary target and stops after penetrating 14.5 cm into the target. What is the average force of the target on the bullet? (1.97 x 104 N) 7. Larry's gravitational potential energy is 1870 J as he sits 2.20 m above the ground in a sky diving airplane. What is Larry's gravitational potential energy when be begins to jump from the airplane at an altitude of 923 m? (784 550 J) 8. A roller-coaster car is moving at 20 m/s along a straight horizontal track. What will its speed be after climbing the 15 m hill shown in the figure if friction is ignored? (10.3 m/s)

9. The initial velocity of a 4.0-kg box is 11 m/s, due west. After the box slides 4.0 m horizontally, its speed is 1.5 m/s. Determine the magnitude and the direction of the non-conservative force acting on the box as it slides. (59.375 N, east) 10. A 9.0-kg box of oranges slides from rest down a frictionless incline from a height of 5.0 m. A constant frictional force, introduced at point A, brings the block to rest at point B, 19 m to the right of point A.

Calculate: (a) The speed of the block just before it reaches point A? (b) The coefficient of kinetic friction, µk, of the surface from A to B? (9.9 m/s, 0.263)

63 | P a g e...