**In this chapter we will introduce the concepts of work and kinetic energy. These tools will significantly simplify the manner in which certain problems can be solved.**

Figure 7.1. A force

**F**acting on a body. The resulting displacement is indicated by the vector**d**.## 7.1. Work: constant force

**Suppose a constant force F acts on a body while the object moves over a distance d. Both the force F and the displacement d are vectors who are not necessarily pointing in the same direction (see Figure 7.1). The work done by the force F on the object as it undergoes a displacement d is defined as**

_{ }

* d = 0: displacement equal to zero

* [phi] = 90deg.: force perpendicular to displacement

Figure 7.2. Positive or Negative Work.

The work done by the force F can be positive or negative, depending on
[phi]. For example, suppose we have an object moving with constant velocity.
At time t = 0 s, a force F is applied. If F is the only force acting on the
body, the object will either increase or decrease its speed depending on
whether or not the velocity v and the force F are pointing in the same
direction (see Figure 7.2). If (**F***

**v**) > 0, the speed of the object will increase and the work done by the force on the object is positive. If (

**F***

**v**) < 0, the speed of the object will decrease and the work done by the force on the object is negative. If (

**F***

**v**) = 0 we are dealing with centripetal motion and the speed of the object remains constant. Note that for the friction force (

**F***

**v**) < 0 (always) and the speed of the object is always reduced !

Per definition, work is a scalar. The unit of work is the Joule (J). From the definition of the work it is clear that:

1 J = 1 N m = 1 kg m

^{2}/s^{2}
Figure 7.3. Forces acting on the safe.

**Sample Problem 7-2**

A safe with mass m is pushed across a tiled floor with constant velocity for a distance d. The coefficient of friction between the bottom of the safe and the floor is u

_{k}. Identify all the forces acting on the safe and calculate the work done by each of them. What is the total work done ?

Figure 7.3 shows all the forces that act on the safe. Since the safe is moving with constant velocity, its acceleration is zero, and the net force acting on it is zero

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_{ }

_{ }

_{ }

_{ }

_{ }

_{ }

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**Example Problem 1**

A crate with mass m is pulled up a slope (angle of inclination is [theta]) with constant velocity. Calculate the amount of work done by the force after the crate has moved to a height h (see Figure 7.4).

Figure 7.4. Example Problem 1.

The coordinate system that will be used is shown in Figure 4. Since
the crate is moving with a constant velocity, the net force in the x and y
direction must be zero. The net force in the x direction is given by_{ }

_{ }

_{ }

_{ }

_{ }

_{ }

Figure 7.5. Crate moved in vertical direction.

If the same crate had been lifted by a height h in the vertical
direction (see Figure 7.5), the force F required to produce a constant velocity
would be equal to
F = m g

This force acts over a distance h, and the work done by this force on
the object is
W

which is equal to the work done by the force on the inclined slope.
Although the work done by each force is the same, the strength of the required
force is very different in each of the two cases._{F}= m g h**Example Problem 2**

A 3.57 kg block is drawn at constant speed 4.06 m along a horizontal floor by a rope exerting a 7.68 N force at angle of 15deg. above the horizontal. Compute (a) the work done by the rope on the block, and (b) the coefficient of kinetic friction between block and floor.

Figure 7.6. Example Problem 2.

A total of four forces act on the mass m: the gravitational force W,
the normal force N, the friction force f_{k}and the applied force F. These four forces are shown schematically in Figure 7.6. Since the velocity of the mass is constant, its acceleration is equal to zero. The x and y-components of the net force acting on the mass are given by

_{ }

_{ }

_{ }

_{k}is given by

_{ }

_{k}is also related to the applies force in the following manner

_{ }

_{ }

_{ }

_{ }

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## 7.2. Work: variable force

**In the previous discussion we have assumed that the force acting on the object is constant (not dependent on position and/or time). However, in many cases this is not a correct assumption. By reducing the size of the displacement (for example by reducing the time interval) we can obtain an interval over which the force is almost constant. The work done over this small interval (dW) can be calculated**

_{ }

_{ }

**Example: The Spring**

An example of a varying force is the force exerted by a spring that is stretched or compressed. Suppose we define our coordinate system such that its origin coincides with the end point of a spring in its relaxed state (see Figure 7.7). The spring is stretched if x > 0 and compressed if x < 0. The force exerted by the spring will attempt to return the spring to its relaxed state:

if x < 0: F > 0

if x > 0: F < 0

It is found experimentally that for many springs the force is
proportional to x:
F = - k x

Figure 7.7. Relaxed, Stretched and Compressed Springs.

where k is the spring constant (which is positive and independent of
x). The SI units for the spring constant is N/m. The larger the spring
constant, the stiffer the spring. The work done by the spring on an object
attached to its end can be calculated if we know the initial position
x_{i}and final position x

_{f}of the object:

_{ }

_{i}= 0) we find that the work done by the spring is

_{ }

Figure 7.8. Pendulum in x-y plane

## 7.3. Work in 2D

**Consider the pendulum shown in Figure 7.8. The pendulum is moved from position 1 to position 2 by a constant force F, pointing in the horizontal direction (see Figure 7.8). The mass of the pendulum is m. What is the work done by the sum of the applied force and the gravitational force to move the pendulum from position 1 to position 2 ?**

**Method 1 - Difficult**

The vector sum of the applied force and the gravitational force is shown in Figure 7.9. The angle between the applied force F and the vector sum F

_{t}is a. Figure 7.9 shows that the following equations relate F to F

_{t}and F

_{g}to F

_{t}:

_{ }

_{ }

Figure 7.9. Vector sum F

In order to calculate the work done by the total force on the
pendulum, we need to know the angle between the total force and the direction
of motion. Figure 7.10 shows that if the angle between the pendulum and the
y-axis is [theta] , the angle between the total force and the direction of
motion is [theta] + a. The distance dr is a function of d[theta]:_{t}of F_{g}and F._{ }

_{t}will not change. The work done by F

_{t}on the pendulum is given by

_{ }

_{t}can be obtained by integrating the equation for dW over all angles between [theta] = 0deg. and [theta] = [theta]

_{max}. The maximum angle can be easily expressed in terms of r and h:

_{ }

_{ }

Figure 7.10. Angle between sum force and direction.

The total work done is_{ }

_{ }

_{t}cos(a), F

_{t}sin(a), r cos([theta]

_{max}) and r sin([theta]

_{max}) we can rewrite this expression and obtain for W:

_{ }

**Method 2 - Easy**

The total work done on the pendulum by the applied force F and the gravitational force F

_{g}could have been obtained much easier if the following relation had been used:

_{ }

_{g}. These two quantities can be calculated easily:

_{ }

_{ }

_{ }

## 7.4. Kinetic Energy

**The observation that an object is moving with a certain velocity indicates that at some time in the past work must have been done on it. Suppose our object has mass m and is moving with velocity v. Its current velocity is the result of a force F. For a given force F we can obtain the acceleration of our object:**

_{ }

_{ }

_{ }

_{ }

_{ }

**kinetic energy K**:

_{ }

_{i}to some final value K

_{f}the amount of work done on the particle is given by

W = K

This indicates that the change in the kinetic energy of a particle is
equal to the total work done on that particle by all the forces that act on
it._{f}- K_{i}**Alternative Derivation**

Consider a particle with mass m moving along the x-axis and acted on by a net force F(x) that points along the x-axis. The work done by the force F on the mass m as the particle moves from its initial position x

_{i}to its final position x

_{f}is

_{ }

_{ }

_{ }

**Example Problem 3**

An object with mass m is at rest at time t = 0. It falls under the influence of gravity through a distance h (see Figure 7.11). What is its velocity at that point ?

Since the object is initially at rest, its initial kinetic energy is zero:

K

The force acting on the object is the gravitational force _{i}= 0 J
F

_{g}= m g
Figure 7.11. Falling Object.

The work done by the gravitational force on the object is simply
W = F

The kinetic energy of the object after falling a distance h can be
calculated:_{g}h = m g h
W = m

and its velocity at that point is^{.}g^{.}h = K_{f}- K_{i}= K_{f}_{ }

Figure 7.12. Projectile motion.

**Example Problem 4**

A baseball is thrown up in the air with an initial velocity v

_{0}(see Figure 7.12). What is the highest point it reaches ?

The initial kinetic energy of the baseball is

_{ }

W = K

In this case the direction of the displacement of the ball is opposite
to the direction of the gravitational force. Suppose the baseball reaches a
height h. At that point the work done on the baseball is_{f}- K_{i}= - K_{i}
W = - m g h

The maximum height h can now be calculated:_{ }

_{ }

## 7.5. Power

**In every day life, the amount of work an apparatus can do is not always important. In general it is more important to know the time within which a certain amount of work can be done. For example: the explosive effect of dynamite is based on its capability to release large amounts of energy in a very short time. The same amount of work could have been done using a small space heater (and having it run for a long time) but the space heater would cause no explosion. The quantity of interest is power. The power tells us something about the rate of doing work. If an amount of work W is carried out in a time interval [Delta]t, the average power for this time-interval is**

_{ }

_{ }

^{.}hour. This is equivalent to

7.3.1. kW

We can also express the power delivered to a body in terms of the force
that acts on the body and its velocity. Thus for a particle moving in one
dimension we obtain^{.}h = (10^{3}W) (3600 s) = 3.6 x 10^{6}J = 3.6 MJ_{ }

_{ }

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