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The work done on an object by an external force is given by the formula

work_done = force

displacement

Work is our first example of a scalar product or dot product. A dot product occurs when two vectors are multiplied together in such a way as to produce a scalar value. Technically, the above definition for work can be calculated with the equation

work_done = ||F|| * ||s|| cos(?)

where

||F|| represents the magnitude or the length of vector F
||s|| represents the magnitude or the length of vector s
? represents the magnitude of the angle between F and s.

Or, it can equivalently be evaluated with the formula

work_done = F_x * s_x + F_y * s_y

where the x- and y-components of F and s are multiplied and then added together.

Let's use an example to show how these two expressions are equivalent.

Suppose a toy cart is sliding between two rails along the surface of a table while being pulled diagonally by a force F.

To determine the work done on the cart by the force, we can use either of these two methods:

work_done = ||F|| * ||s|| cos(?)
work_done = F_x * s_x + F_y * s_y

Method #1: work_done = ||F|| * ||s|| cos(?)

||F|| = 10 N
||s|| = 2 m

? = 37º + | -23º | = 60º

work_done = ||F|| * ||s|| cos(?) = (10)(2) cos (60º) = 10 J

Notice that this result could alternatively be calculated with the expression:

work_done = [F cos(?)] s = [10 cos(60º)](2) = 10 J

where F cos(?) is the component of F in the direction of s.

Method #2: work_done = F_x * s_x + F_y * s_y

F_x = F cos(?) = 10 cos(37º) = 8 N
F_y = F sin(?) = 10 sin(37º) = 6 N
s_x = s cos(?) = 2 cos(-23º) = 1.84 m
s_y = s sin(?) = 2 sin(-23º) = -0.781 m

work_done = F_x * s_x + F_y * s_y = (8)(1.84) + (6)(-0.781) = 10 J

Any slight numerical differences between the two calculations would be the result of rounding decimal expressions.

The formula that is usually used to calculate the amount of work done on a mass M by a constant force F acting at an angle ? to its displacement s is W = Fs cos(?)

Using this formula when ? = 90^o the work_done = 0 since cos(90^o) = 0. Referring to the above diagram, the vertical component, F sin(?), of F would not result in any work being done on mass M since it acts at right angles to the displacement s.

Work done on an object by a force acting parallel to its displacement can be expressed as simply (where ? = 0º)

work_done = Fs

Work can also be calculated as W_done = ?F ds or the equivalent area under a force vs displacement graph.

Conservation of Energy and Springs

When a spring is compressed, work is done on the spring by the external agent exerting the force. Suppose that this work is done by a moving object which strikes and sticks to a spring initially in its equilibrium position [position A] so that as the moving object loses kinetic energy (eventually coming to rest) it does work on the spring to compress it [position B]. We will not at this time continue our examination into when the spring later rebounds.

Numerically we would therefore set the magnitude of the kinetic energy lost by the object equal to the elastic potential energy gained by the spring. This will allow us to either solve for the maximum compression distance in the spring; or vice versa, if given the compression produced in the spring, solve for the original velocity of the colliding mass.

Horizontally Oscillating Springs

Now suppose that a spring is initially compressed and then released on a frictionless surface. It then oscillates with one end firmly attached to a base of support and a mass attached to its free end. As the mass vibrates back and forth, the energy in the system transforms between PE_e (at the endpoints of the oscillation) and KE (as the mass passes through equilibrium). Since the surface is frictionless no mechanical energy is lost to thermal energy as the mass slides back and forth over the surface.

The maximum kinetic energy occurs as the mass passes through equilibrium.

When a spring has been set into oscillatory motion, the equation used to calculate its period is

Remember that frequency and period are reciprocals: f = 1/T. The unit to measure frequency is hertz, or vibrations per second; while the unit to measure period is seconds, or seconds per vibration.

Energy Conservation in Simple Pendulums

Remember the formula used to calculate the gravitational potential energy of a mass given its mass and height above an arbitrary zero level is

PE_gravity = mgh

When a pendulum is pulled back from equilibrium through an angle ?, its height is calculated with the formula

h = L - L cos ?
where ? is the angular displacement

The formula used to calculate the kinetic energy of a massive particle is

KE = ½ mv²

In the absence of non-conservative forces, such as friction or applied, external forces, the mechanical energy in a system is conserved. That is

Mechanical Energy

Energy is defined as the ability to do work and is a scalar quantity. Energy has no direction, only magnitude. Mechanical energy comes in two varieties: kinetic energy (KE) and potential energy (PE).

Kinetic Energy

Kinetic energy represents the energy caused by an object's motion

KE = ½mv²

where v is the object's actual speed, that is, the magnitude of the object's instantaneous resultant velocity. In this formula, m must be measured in kg and v must be measured in m/sec. Note that this collection of units

kg (m/sec)² = kg m²/sec² is called a joule (J).

Momentum and Energy

The relationship between conservation of energy and conservation of momentum is an extremely important one. During every collision, momentum is conserved. Remember that conservation of momentum is actually a restatement of Newton's Third Law.

Occasionally kinetic energy is also conserved during an elastic collision. When this type of collision is being references problems will explicitly state that a steel ball or a "superball" bounces elastically off a surface. The implication is that the ball will return to the same height from which it is released. That is, its original gravitational potential energy is transformed into kinetic energy which remains constant throughout the collision allowing the ball to return to its original height. Those collisions are rare and far between.

But even when collisions lose KE, and are considered to be inelastic, we often use conservation of energy methods to analyze the behaviors of the objects involved. We do this by considering the energy content of the system before the objects collide and then again after they collide - just NOT during the collision

Potential Energy Functions

the use of the term mechanical energy usually involves three types of energies: potential gravitational energy, kinetic energy, and elastic potential energy. Although potential energy is often represented by the expression PE, in this lesson we will use the variable U; similarly, kinetic energy will be represented by the variable K.

In the absence of non-conservative, or dissipative forces, these energies obey the law of conservation of energy, or ?U + ?K = 0. That is, when a system is only acting under the influence of conservative forces its total energy content never changes, the energy just converts between forms.

Force and Position Relationships

To look into this more carefully, let's re-examine some important graphs for a vibrating spring.

Notice that the position and acceleration/force graphs are 180º out-of-phase: when the spring's displacement from its equilibrium is UP, the restoring force and acceleration are DOWN and vice-versa.

Our next graphs draw our attention to the spring's displacement, energy modes and restoring force.

Notice that

when the spring is either in a state of maximum extension or compression its potential energy is also a maximum
when the spring's displacement is DOWN the restoring force is UP
when the potential energy function has a negative slope, the restoring force is positive and vice-versa
when the restoring force is zero, the potential energy is zero
at any point in the cycle, the total energy is constant, U + K = U_max = K_max

Force Functions

Our next step will be to show that a function representing the instantaneous values of the restoring force can be expressed as the negative of the derivative of our potential energy function

Remember our two relationships involving work

The work done by a conservative force decreases an object's potential energy while it is increasing its kinetic energy

Defining the initial potential energy U_o = 0, gives us

Using the calculus, we see that our desired expression of the instantaneous restoring force being equal to the negative derivative of the potential energy function.

Rotational Dynamics: Pivoting Rods

Suppose one end of a uniform rod is pivoted against a wall and the other end is suspended by a rope from the ceiling. While it is in equilibrium, the question of what force the hinge supplies is a reasonably simple task.


net F_parallel = 0		not applicable (there are no horizontal forces)
net F_{perpendicular} = 0		= mg + T
net ? = 0		mg(L/2) = T(L) which reduces to T = mg/2
Substituting the value for tension found in solving net ? = 0 into the equation for net perpendicular force shows that the only non-zero component of the force at the hinge must also equal ½mg.

Horizontal Rotating Rod

What vertical force will the hinge be required to supply at the instant just after the string is cut? Will its upward support remain mg/2? or would it be greater? or perhaps smaller? To work this problem we once again look at the same equations, but this time from the perspective of accelerated motion.


F_parallel		not applicable (the rod has no instantaneous angular velocity)
net F_{perpendicular} = ma_tangential		mg - = ma_tangential
net ? = I?		mg(L/2) = ?mL²? which reduces to ? = 3g/2L
Substituting ? back into the equation for net F_{perpendicular} = ma_tangential mg - F_? = mr? mg - F_? = m(L/2)(3g/2L) F_? = mg - (3/4)mg F_? = ¼mg

Pivoting Rod

But how does the force supplied by the hinge change as the rod continues to rotate? Let's examine what happens at an instantaneous angle ? which is formed between the wall and the rotating rod.

First we need to notice that the center of gravity of the rod is moving closer to the wall and is sweeping out an arc as the rod rotates. Later when we consider conservation of energy, we will recall this behavior when we state that the center of gravity's potential energy falls through a height of L/2.

Notice that our freebody diagram has "pivoted" with the rod. That is, we are no longer concerned with the customary "vertical and horizontal" forces on the hinge since the rod has an angular acceleration. Now we are interested in the forces that act perpendicular to and parallel with the rod.

As with anything in circular motion, every point on the rod, in particular, the rod's center of mass, is experiencing a centripetal acceleration.

To calculate the angular velocity of the rod, we must use conservation of energy techniques since the rod's angular acceleration is not uniform. The change in the potential energy of the rod is basically a calculation of the vertical displacement of the rod's center of gravity.

Our statement of conservation of energy is

Before continuing with our calculations for the forces at the hinge, we need to re-examine our freebody diagram a little more thoroughly. Not only are there parallel and perpendicular components to the force on the hinge, the weight also has components that are parallel and perpendicular to the rod.

Returning to our calculation for F_parallel, we can now write an expression for the centripetal force, or the net force to the center of rotation.

To calculate F_{perpendicular}, we will use Newton's 2nd Law in both its translational and rotational forms.

The magnitude of the resultant force on the hinge can now be calculated using the Pythagorean Theorem.

Notice that, when our pivoting rod is released from a horizontal position, the magnitude of the net force on the hinge is independent of the rod's instantaneous angle!

Tension Cases: Four Special Situations

Case I: Static Equilibrium - There is no acceleration in either the x- or y-dimension. In each case, ?F = 0.

?F_x = 0 ? T_cord + (-T sin ?) = 0
?F_y = 0 ? T cos ? - mg = 0

Case II: Conical Pendulum - The forces are balanced in the y-dimension but there is an unbalanced x-dimension force directed towards the center of the circle.

?F_x = ma_c ? T sin ? = ma_c
?F_y = 0 ? T cos ? - mg = 0

Case III: Vertical Circles - In vertical circular motion, the acceleration is not uniform as gravity speeds up objects while they fall and slows them down as they rise. Tension is greatest at the bottom of a vertical circle and approach minimum values while passing through the top of a vertical circle.

top

bottom


top		bottom
	net F_c = T + mg		net F_c = T - mg
	m(v²/r) = T + mg		m(v²/r) = T - mg
	T = m(v²/r) - mg		T = m(v²/r) + mg

The following formula used to calculate the minimum, or critical, speed required for the block to pass through the top of a vertical circle is derived by taking the limit as T ? 0 in the previous formula for centripetal force at the top of a vertical circle and solving for v:

m(v²/r) = mg
v²/r = g
v² = rg
v_critical = ?(rg)

Case IV: Simple Pendulum - As a simple pendulum swings, its bob experiences a tangential acceleration along its arc and a centripetal acceleration towards the center of the circle. Remember that a pendulum is merely the bottom of a vertical circle. Energy methods should be used to calculate the velocity at ang given time. To review this procedure, visit this related lesson on energy conservation in simple pendulum

?F_x = ma_t ? mg sin ? = ma_t
?F_y = ma_c ? T - mg cos ? = ma_c

Work

In order for an object to gain energy, work must be done on it by an external force. When work is done on an object by a force acting parallel to its displacement the formula is:

work_done = force x displacement

For positive work to be done, F and s must be parallel and pointed in the same direction. The unit used to measure work and energy is a joule. [J = kg m²/sec² = Nm]

Work done by non-conservative forces

If we look at the forces on an object being pulled across a table's surface there would be three: F, the applied force, N, the normal or supporting force supplied by the table, and mg, its weight or the gravitational force of attraction to the earth.

The normal force and the object's weight are in static equilibrium (they are balanced forces), the applied force, F, is an unbalanced force and will result in the object being accelerated across the top of the table's surface in the same direction as the force. This acceleration will change the object's velocity and subsequently its kinetic energy. We say that this applied force is doing work on the object. The amount of work done by F is directly proportional to the distance through which the force is applied as it pulls the object across the table's surface.

net Work_done = (net F)s

By using Newton's second law, net F = ma, our equation becomes

net Work_done = (ma)s
net Work_done = m(as)

Remembering the kinematics equation v_f² = v_o² + 2as and solving for "as" let's our equation become

net Work_done = m[½(v_f² - v_o²)]
net Work_done= ½(mv_f² - mv_o²)
net Work_done= ½mv_f² - ½mv_o²
net W_done= ?KE

The relationship we just derived is called the energy-work theorem..

This statement tells us that when an external force does work on an object it will change the object's kinetic energy; that is, it will cause the object to either gain or lose speed. When more than one force is acting on an object, all forces that are either parallel or antiparallel to the direction the object moves will do work. If the object's velocity remains constant, that just means that the work done by opposing forces (for example, a forward applied force, F, and an opposing force, friction) are equal.

Note that if F and s are perpendicular to each other no work is done on the object. In our example of the block being dragged across the table, neither the normal force nor the weight would do any work on the block since they act at right angles to the direction of the block's motion. Another example would be when a satellite is being held in circular orbit by the force of gravity. Note that since the satellite's speed and orbital radius remain constant, no energy is being changed; therefore, no work is being done on the satellite.

Work done by conservative forces

Work done by conservative forces, or path-independent forces, results in changes in the object's potential energy.

Let's use gravity an example of a CONSERVATIVE FORCE (or path-independent force). Remember that the changes in an object's potential energy only depend on comparing its starting position and its ending position, not on whether it does or does not pass through various points in-between. The block's final change in potential energy is the same whether it follows the path with the intermediate stops B, C and D or whether it is directly taken from A to E. The height of the post is the same.

When you observe an object falling, it loses potential energy (height) while it gains kinetic energy (speed). That is, in the absence of another external, non-conservative force, such as friction, pushing/pulling, or tensions in strings, the total amount of potential energy before the fall equals the total amount of kinetic energy after the fall and the energy-work theorem is restated as the Law of Conservation of Energy:

Work_done = ?KE
Work_{done conservative force}= - ?PE

- ?PE = ?KE
- (PE_f - PE_o) = KE_f - KE_o
- PE_f + PE_o = KE_f - KE_o

KE_o+ PE_o = KE_f + PE_f

For projectiles in freefall this statement of conservation of energy can be used to compare the energies at two different locations (A and B) in its trajectory:

PE_A + KE_A = PE_B + KE_B

Power

The formula used to calculate the power delivered or developed to complete a specified amount of work in a given time is

Power = work/time

Power is measured in watts. [watt = J/sec]

A second form of power is easily derived.

Power = work/time
Power = (Fs) / t
Power = F (s/t)
Power = Fv

This expression can only be used when either the velocity is constant or when only an instantaneous value for power is being requested.

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