Friction Involved

We next consider the example in Fig. 8-13a. A constant horizontal force pulls a block along an axis and through a displacement of magnitude , increasing the block's velocity from to . During the motion, a constant kinetic frictional force from the floor acts on the block. Let us first choose the block as our system and apply Newton's second law to it. We can write that law for components along the axis ( ) as

Because the forces are constant, the acceleration is also constant. Thus, we can use Eq. 2-16 to write

.

Solving this equation for , substituting the result into Eq. 8-25, and rearranging then give us

because for the block,

In a more general situation (say, one in which the block is moving up a ramp), there can be a change in potential energy. To include such a possible change, we generalize Eq. 8-27 by writing

8-28

By experiment we find that the block and the portion of the floor along which it slides become warmer as the block slides. As we shall discuss in Chapter 19, the temperature of an object is related to the object's thermal energy (the energy associated with the random motion of the atoms and molecules in the object). Here, the thermal energy of the block and floor increase because (1) there is friction between them and (2) there is sliding. Recall that friction is due to the cold-welding between two surfaces. As the block slides over the floor, the sliding causes repeated tearing and reforming of the welds between the block and the floor, which makes the block and floor warmer. Thus, the sliding increases their thermal energy .

Through experiment, we find that the increase in thermal energy is equal to the product of the magnitudes and :

(-increase in thermal energy by sliding). 8-29

Thus, we can rewrite Eq. 8-28 as

8-30

is the work done by the external force (the energy transferred by the force), but on which system is the work done (where are the energy transfers made)? To answer, we check to see which energies change. The block's mechanical energy changes, and the thermal energies of the block and floor also change. Therefore, the work done by force F is done on the block-floor system. That work is

(8-31)

This equation, which is represented in Fig. 8-13b, is the energy statement for the work done on a system by an external force when friction is involved.

8-7 Conservation of Energy

We now have discussed several situations in which energy is transferred to or from objects and systems, much like money is transferred between accounts. In each situation we assume that the energy that was involved could always be accounted for; that is, energy could not magically appear or disappear. In more formal language, we assumed (correctly) that energy obeys a law called the law of conservation of energy, which is concerned with the total energy of a system. That total is the sum of the system's mechanical energy, thermal energy, and any form of internal energy in addition to thermal energy. (We have not yet discussed other forms of internal energy.) The law states that

> The total energy of a system can change only by amounts of energy that are transferred to or from the system.

The only type of energy transfer that we have considered is work done on a system. Thus, for us at this point, this law states that

where is any change in the mechanical energy of the system, is any change in the thermal energy of the system, and is any change in any other form of internal energy of the system. Included in are changes in kinetic energy and changes in potential energy (elastic, gravitational, or any other form we might find).

This law of conservation of energy is not something we have derived from basic physics principles. Rather, it is a law based on countless experiments. Scientists and engineers have never found an exception to it.

Isolated System

If a system is isolated from its environment, then there can be no energy transfers to or from it. For that case, the law of conservation of energy states:

The total energy of an isolated system cannot change.

Many energy transfers may be going on within an isolated system, between, say, kinetic energy and a potential energy or kinetic energy and thermal energy. However, the total of all the forms of energy in the system cannot change.

We can use the rock climber in Fig. 8-14 as an example, approximating her, her gear, and Earth as an isolated system. As she rappels down the rock face, changing the configuration of the system, she needs to control the transfer of energy from the gravitational potential energy of the system. (That energy cannot just disappear.) Some of it is transferred to her kinetic energy. However, she obviously does not want very much transferred to that form or she will be moving too quickly, so she has wrapped the rope around metal rings to produce friction between the rope and the rings as she moves down. The sliding of the rings on the rope then transfers the gravitational potential energy of the system to thermal energy of the rings and rope in a way that she can control. The total energy of the climber-gear-Earth system (the total of its gravitational potential energy, kinetic energy, and thermal energy) does not change during her descent.

For an isolated system, the law of conservation of energy can be written in two ways. First, by setting in Eq. 8-33, we get

We can also let , where the subscripts 1 and 2 refer to two different instants, say before and after a certain process has occurred. Then Eq. 8-34 becomes

Equation 8-35 tells us:

In an isolated system, we can relate the total energy at one instant to the total energy at another instant without considering the energies at intermediate times.

This fact can be a very powerful tool in solving problems about isolated systems when you need to relate energies of a system before and after a certain process occurs in the system.

In Section 8-4, we discussed a special situation for isolated systems—namely, the situation in which nonconservative forces (such as a- kinetic frictional force) do not act within them. In that special situation, and are both zero, and so Eq. 8-35 reduces to Eq. 8-18. In other words, the mechanical energy of an isolated system is conserved when nonconservative forces do not act within it.

Power

Now that you have seen how energy can be transferred from one form to another, we can expand the definition of power given in Section 7-7. There it is the rate at which work is done by a force. In a more general sense, power is the rate at which energy is transferred by a force from one form to another. If an amount of energy is transferred in an amount of time , the average power due to the force is

Similarly, the instantaneous power due to the force is

(8-36)

Similarly, the instantaneous power due to the force is

(8-37)

Sample Problem 8-7

In Fig. 8-15, a 2.0 kg package of tamale slides along a floor with speed v₁ = 4.0 m/s. It then runs into and compresses a spring, until the package momentarily stops. Its path to the initially relaxed spring is frictionless, but as it compresses the spring, a kinetic frictional force from the floor, of magnitude 15 N, acts on it. The spring constant is 10,000 N/m. By what distance d is the spring compressed when the package stops?

SOLUTION: A starting Key Idea is to examine all the forces acting on the package, and then to determine whether we have an isolated system or a system on which an external force is doing work.

Forces: The normal force on the package from the floor does no work on the package, because its direction is always perpendicular to that of the package's displacement. For the same reason, the gravitational force on the package does no work. As the spring is compressed, however, a spring force does work on the package, transferring energy to elastic potential energy of the spring. The spring force also pushes against a rigid wall. Because there is friction between the package and the floor, the sliding of the package across the floor increases their thermal energies.

System: The package-spring-floor-wall system includes all these forces and energy transfers in one isolated system. Therefore, a second Key Idea is that, because the system is isolated, its total energy cannot change. We can then apply the law of conservation of energy in the form of Eq. 8-35 to the system:

(8-38)

Let subscript 1 correspond to the initial state of the sliding package, and subscript 2 correspond to the state in which the package is momentarily stopped and the spring is compressed by distance . For both states the mechanical energy of the system is the sum of the package's kinetic energy ( ) and the spring's potential energy ( ). For state 1, (because the spring is not compressed), and the package's speed is |. Thus, we have

For state 2, (because the package is stopped), and the compression distance is . Therefore, we have

Finally, by Eq. 8-29, we can substitute for the change in the thermal energy of the package and the floor. We can now re­write Eq. 8-38 as

.

Rearranging and substituting known data give us

.

Solving this quadratic equation yields

d = 0.055 m = 5.5 cm.

(Answer)

EXAMPLE 7—5 A man holds a ball of mass kg at rest in his hand. He then throws the ball vertically upward. In this process, his hand moves up 0.5 m before the ball leaves his hand with an upward velocity of 20 m/s. Discuss the motion of the ball from the work-energy standpoint, assuming m/s².

SOLUTION First, consider the throwing process. Take the reference level ( ) at the initial position of the ball. Then , . Take point 2 at the point where the ball leaves the thrower's hand. Then

= (0.2 kg)(10 m/s²)(0.5 m) = 1.0 J,

= 1(0.2 kg)(20 m/s)² = 40 J.

Let represent the upward force exerted on the ball by the man in the throwing process. The work is then the work done by this force and is equal to the sum of the changes in kinetic and potential energy of the ball. The kinetic energy of the ball increases by 40 J and its potential energy by 1 j. The work done by the upward force is therefore 41 J.

If the force is constant, the work done by this force is given by

and the force is then

However, the work done by the force is 41 J whether the force is constant or not.

Now consider the flight of the ball after it leaves the thrower's hand. In the absence of air resistance, the only force of the ball is then its weight . Hence the total mechanical energy of the ball remains constant. The calculations are simpler if we take a new reference level at the point where the ball leaves the thrower's hand. Calling this point 1, we have

J, ,

,

and the total mechanical energy at any point in the path is 40 J.

Suppose we want to find the speed of the ball at a height of 15 m above the reference level. Its potential energy at this elevation is

Therefore the kinetic energy at this point is J, and the speed at this point is given by

, m/s.

The significance of the ± sign is that the ball passes this point twice, once on the way up and again on the way down. Its potential energy at this point is the same whether it is moving up or down. Hence its kinetic energy is the same and its speed is the same. The algebraic sign of the velocity is + when the ball is moving up and — when it is moving down.

Next, let us find the highest point the ball reaches. At this point and . Therefore at this point ], and the maximum height of the ball above the thrower's hand is given by

J

And m.

Finally, suppose we are asked to find the ball's speed at a point 30 m above the reference level. The potential energy at this point would be 60 J. But the total energy is only 40 J, so the ball can never reach a height of 30 m.

EXAMPLE 7-6 A child slides down a curved playground slide that is one quadrant of a circle of radius , as in Fig. 7-9. If he starts from rest and there is no friction, find his speed at the bottom of the slide. (The motion of this child is exactly the same as that of a child on a swing of length , with the other end held at point O.)

SOLUTION We cannot use the equations of motion with constant acceleration because the acceleration decreases during the motion. (The slope angle of the slide becomes smaller and smaller as the body descends.) If there is no friction, however, the only force on the child other than his weight is the normal force exerted on him by the slide. The work done by this force is zero because at each point it is perpendicular to the small element of displacement near that point. Thus and mechanical energy is conserved. Take point 1 at the starting point and point 2 at the bottom of the slide. Take the reference level at point 2. Then , , and

,

The speed is therefore the same as if the child had fallen vertically through a height . (What is now the significance of the ± sign?) As a numerical example, let = 3.00 m. Then

m/s.

EXAMPLE 7—7 Suppose a child of mass 25.0 kg slides down a slide of radius R = 3.00 m, like that in Fig. 7-9, but his speed at the bottom is only 3.00 m-s^-1. What work was done by the frictional force acting on the child?

SOLUTION In this case,

J

The frictional work was therefore -623 J, and the total mechanical energy decreased by 623 J. The mechanical energy of a body is not conserved when friction forces act on it.

EXAMPLE 7—9 A child of weight sits on a swing of length , as shown in Fig. 7—11. A variable horizontal force that starts at zero and gradually increases is used to pull the child very slowly (so the kinetic energy is negligibly small) until the swing makes an angle with the vertical. Calculate the work done by the force .

SOLUTION The sum of the works done by all the forces other than the gravitational force must equal the change of total energy, that is, the change of

kinetic energy plus the change of gravitational potential energy. Hence

Since is perpendicular to the path at its point of application, ; and since the swing was pulled very slowly at all times, the change of kinetic energy is also zero. Hence

,

where is the distance that the child has been raised. From Fig. 7-11, .Therefore

.

Note that this result agrees with that obtained in Example 7-3, where W_P was calculated directly.

Find the work done in moving a particleali

Q. 1.13. Find the work done in moving a particle alone a vector meters, if the applied force is Newton.

Solution.

Here, J

Derive an expression for the gravitational potential energy of a body lying at a height ( ), radius of earth) above the surface of earth.

Define kinetic energy. Derive an expression for the kinetic energy of a body moving with a uniform velocity.

Derive the expression for the kinetic energy of a mass m moving with a velocity v. Define SI unit of energy.

State and prove work-energy theorem.

Define work, power and energy giving their units in SI. Prove that work done by constant force is equal to the total change in kinetic energy of the body, whose initial and final velocities are и and v respectively.

Explain the meaning of the term work. Calculate the work done by a constant force. Is work a scalar or a vector quantity ?

Explain the term work. Show that work done is equal to dot product of force and displacement.

What is meant by positive work, negative work and zero

work? Give one example of each.

Define work. What is SI unit of work ? What is meant by positive work, negative work and zero work ?

Explain, how can we find the work done by a variable force.

What is a conservative force ? Explain its various properties.

Define power. What is SI unit of power ? Prove that instantaneous power is given by the scalar product of force and velocity.

Define and explain the terms : work, energy and power. State their SI units.

Define energy and power. What are their units in SI

system?

Define work and power. State their SI units.

Show that the total mechanical energy of a body f ailing freely under gravity is conserved.

What are elastic and inelastic collisions? Give examples.

Type A. On Work done

1. What is the work done by a man in carrying a suitcase weighing 30 kg over his head, when he travels a distance of 10 m in the (/) vertical direction, (ii) horizontal direction ?

[Ans. (z) 2,940 J; (ii) zero]

2. Find the work done, if a weight of 25 kg is lifted through a vertical height of 2 m from the ground and also if it is raised to same place by pushing up an inclined plane making an angle of 30° with the ground. Neglect friction.

[Ans. 490 J; 490 J]

3. A man weighing 50 kg climbs 10 m. Calculate the work
done by gravity. [Ans. - 4,900 J]

4. A particle is displaced through meters under the influence of a force newton. Calculate the work done.

( [Ans. - 15 J]

5. A particle moves from a point to position under the action of a force newton. If the displacement between two points is in meters, calculate the work done.

[Ans. 46 J]

On Potential and Kinetic energy

6. 230 joule were spent in lifting a 10 kg weight to a height of 2 m. Calculate the acceleration with which it was raised. Take g = 10 m/s². [Ans. 1.5 m/s]

7. Calculate the work done in lifting a 300 N weight to a height of 10 m with an acceleration 0.5 m/s². Take g = 10 m/s² [Ans. 3,150 J]

8. A body of mass 2 kg initially at rest is moved by a horizontal force of 0-5 N on a smooth (frictionless) table. Obtain the work done by the force in 8 s and show that this equals the change in kinetic energy of the body.

[Ans. 4 J]

9. A block of mass 2 kg is lying on the frictionless table. A force of 8 N is applied on it for 12 s. Calculate its kinetic energy. [Ans. 2,304 J]

Date: 2015-01-12; view: 1430