Friction

The force of static friction does no work and you do no work against this force.  Even though you may be pushing hard against an object acted on by the force of static friction, the object is not moving.  The displacement is zero and W = 0.

You, however, do work against the force of sliding friction.  If you push with force F on a cabinet, while the cabinet moves with constant velocity through a distance d in the direction of F, the work you do on the cabinet is W = Fd = mNd.  The force of fiction does negative work W = -mNd on the cabinet.

Work is the conversion of one form of energy into another.  You do positive work on the cabinet, but the cabinet is not gaining potential energy, (it is not being pushed up a hill), and it is not gaining kinetic energy (it is moving with constant speed).  The work done by the frictional force (negative work) on the cabinet transforms the energy you transfer to the object into disordered energy.

Sliding friction converts most of the work you do on the file cabinet into thermal energy.  As you slide the cabinet across the floor, the bottom of the cabinet and the floor get warmer.  This thermal energy cannot easily be converted back into ordered energy, it cannot easily be used to do useful work.

Sliding friction also causes wear.  Some of the work you do on the file cabinet is converted into electrostatic potential energy.  The atoms and molecules in a solid object are held together by electrical forces, and you have to do work to break them apart.  To detach a cluster of atoms from the bottom of the file cabinet you must do work against the electrostatic force, in the same way you must do work against the gravitational force to lift a rock up from the ground.  As you slide your cabinet across the floor, some material is removed from the bottom of the cabinet and from the floor.  It will not take long before damage becomes visible.

A cart on wheels can roll across the floor.  Rolling involves only static friction, not sliding friction. Static friction does no work.  Rolling does not involve the production of thermal energy.  However, to attach two wheels to the cart, you need an axle.  A smooth bearing lets the axle slide inside the hub of the wheel.  The force of sliding friction converts some of the kinetic energy into heat, and exerts a torque on the wheel.  However, the surfaces of the bearing are usually very smooth and the hub has a small radius, so that the total amount of work done against frictional forces is small.  Adequate lubrication can reduce it even further.

In the diagram above the force of static friction prevents the contact point between wheel and ground from slipping as the applied force accelerates the cart.  The applied force does positive work and the cart gains kinetic energy.

You get in your car, start the engine, put it into "drive", step on the accelerator, and accelerate.
What makes the car start moving forward?

What is missing?

How does friction accelerate your car?

Assume you are dropping a ball onto a perfectly hard floor.  It will rebound, but even the "liveliest " ball will not rise back to its starting position.
Why does it bounce back, and why does it loose height on the rebound?

The ball behaves like a spherical spring.  When the ball hits the floor it exerts a force on the floor larger than its weight, and the floor exerts a force on the ball of equal magnitude.  This force and the gravitational force acting on the ball compress the ball.  As long as the compression is small, Hooke's law is satisfied, the force compressing the ball is proportional to the displacement of the ball from its equilibrium shape.  The gravitational potential energy the ball has before it is dropped is converted into kinetic energy while the ball is falling and then into elastic potential energy when the ball is compressed.  But because the material the ball is made of is not perfectly elastic, internal friction converts some of the energy into thermal energy.

The elastic potential energy stored in the ball when it has lost all its kinetic energy is converted back into kinetic and gravitational potential energy.  The thermal energy, however, is not converted back.

The ball on the floor acts like a compressed spring.  It pushes on the floor with a force proportional to its displacement from its equilibrium shape.  The floor pushes back with a force of equal magnitude in the upward direction.  This force is greater in magnitude than the weight of the ball.  The net force is in the upward direction and the ball accelerates upward.  When the ball's shape is the shape it has when it is sitting still on the floor, (just slightly squashed), the net force is zero.  When its shape relaxes further, the net force is in the downward direction.  But it already has velocity in the upward direction, so it keeps on going upward until its speed has decreased to zero.  Because some of its initial gravitational potential energy has been converted into thermal energy it does not regain its initial height.

A particular ball is characterized by its coefficient of restitution, the ratio of its rebound speed to its collision speed just above the surface of the perfectly hard floor.  A perfectly hard floor is a floor that does not move itself.

coefficient of restitution = (outgoing speed)/(incoming speed)

The coefficient of restitution depends on the material the ball is made of and is always smaller than 1.  The coefficient of restitution is a ratio of speeds.  To find the ration of the outgoing to the incoming kinetic energy we use

K_out/K_in = v²_out/v²_in = (coefficient of restitution)².

Real surfaces are not perfectly hard.  They distort when hit by the ball.  They store energy themselves, and return some of it to the ball as it rebounds.  "Lively" surfaces, such as a trampoline, store energy very efficiently and return almost all of it to the rebounding object.

Problem:

You just got your new car.   You want to experience it.  You want to "feel" its power.  You floor the gas pedal, and you experience a force pressing you back into your seat.  Where does this force come from?

This force is a fictitious force.  Fictitious forces appear in accelerating reference frames.  Such frames are not inertial reference frames.  The accelerating reference frame in the above example is accelerating with your car.  In this frame your car is at rest.  You are at rest in the car, but a force is pushing you against the back of your seat.  To your friend observing you from the sidewalk things look different.  The motor is responsible for the forward acceleration of the car.  Because of your mass, you have inertia.  Without a force acting on you, you would remain at rest.  To keep you accelerating forward, the back of the seat has to push on you.  (You will be pushing on the seat with a force equal in magnitude, but opposite in direction.)  The real force acting on you is in the forward direction, while the fictitious force experienced in the accelerating frame is in the backward direction.

In the accelerating frame of the car, you experience the fictitious force in the backward direction and your weight, pointing down.  The net force experienced is the vector sum of these two forces.  This force becomes your apparent weight, which points in a direction backward and down.

The apparent weight of an object is the vector sum of its real weight and the negative of all the forces that produce the acceleration a = dv/dt

w_apparent = w_real - ma.

Assume you are riding on a merry-go-round.  A reference frame in which you are stationary, i.e. a frame that is moving with you as you are moving along a circular path, is also an accelerating frame.  This frame is moving with constant speed, but the direction of its velocity is constantly changing.  You are sitting still on your seat while the merry-go-round is turning.  But something seems to be pulling you towards the outside, away from the center.  You experience a fictitious force, called the centrifugal force.  To your friend on the ground things again look different.  You are moving in a circle.  The direction of your velocity is constantly changing.  You are accelerating.  The direction of your acceleration is towards the center of the circle, so there must be a force pushing or pulling you toward the center.  If you are sitting in a seat, the wall of the seat will be pushing against you, pushing you towards the center.

As an object moves along a circular path, the direction of its velocity is constantly changing.  The direction of the acceleration is towards the center of the circle.  The magnitude of the acceleration, v²/r, depends on the speed and the radius of the circle along which the object traveling.   In an accelerating frame, which is moving with the object, the object is at rest.  But it experiences the fictitious centrifugal force, constantly trying to throw it towards the outside of the circle.

In many Science Fiction books, humans live in space in a space station that is rotating about a central axis.  In the accelerating frame of the space station, they experience the fictitious centrifugal force, pressing them against the outer wall of the space station.  If this wall is the floor of some room without windows, and the rotation speed and the radius of the station are adjusted so that a = v²/r = g, then there is no way a human or a scientific instrument in the room can distinguish between the centrifugal force and the force of gravity.  If the human steps on a scale, the scale will read the same "weight" as it does on the surface on earth.  If the human throws a ball near the "surface", the ball will follow the same trajectory it would on earth.

In an amusement park the most thrilling rides let the rider experience the largest fictitious forces.  Fictitious forces are experienced in accelerating reference frames.  The seat or cart in which the rider is sitting is accelerating, because it is speeding up, slowing down, or turning rapidly.  A ride that is just moving you along with constant velocity will probably not excite you.  You might as well be in your car on the freeway with the cruise control engaged.

Roller coasters offer thrilling rides.  Most roller coaster rides start with a steep "vertical" drop.  The cart descends on a track down a very steep slope.

Assume the top of the hill is 50m above the ground.  The gravitational potential energy of a object on top of the hill is Mgh.  Neglecting friction, this potential energy has been converted into kinetic energy when the object reaches the bottom of the hill.
K = (1/2)Mv²=  Mgh.
v² = 2gh = 2´9.8 m/s²´50m = 980(m/s)².
v = 31.1m/s = 70mph.
At the bottom of the hill the object (the cart, you, …) moves with a speed of 70 mph.

The forces on the object as it descends are the force of gravity and the support from the track.  The component of the gravitational force perpendicular to the track is canceled by the support from the track, and the component of the gravitational force tangential to the track accelerate the object. However if you are sitting in the accelerating cart, then you are experiencing a fictitious force in the direction opposite to the direction of the velocity in addition to the other forces. This force cancels the tangential component of gravity.  You are left with an apparent weight only due to the perpendicular component of the force of gravity, much less than your normal weight.

If the track pointed straight down, the perpendicular component of the gravitational force would be zero, and your apparent weight would be zero.  You would feel weightless.  A friend on the ground, however, observes you accelerating with acceleration g because of your weight.  The weight of the cart also accelerates the cart at 9.8 m/s².  Your velocity and the velocity of the cart change at the same rate, your relative velocity stays zero.  You are both in free fall, not subject to any forces other than gravity, just as a skydiver, who stepped out of an airplane.

Since the cart of a roller coaster is attached to a track, it is possible, using motors, to produce accelerations in the direction of the velocity greater than g.

Many roller coasters have a vertical loop, 20 m high.  Assume a cart enters the loop at a speed of 25m/s or 56mph. As the cart moves through the loop, it is moving in a circle and therefore it is accelerating.  An observer on the ground concludes that the forces on the cart are the force of gravity and the support from the track.  The support from the track has to cancel the component of the weight perpendicular to the track.  But if it just did that, the cart would keep on moving tangential to the track, since there would be no force pointing towards the center of the circle, providing the centripetal acceleration.  So the force that the track exerts on the cart must be the sum of the perpendicular component of mg and the centripetal force mv²/r.

A rider in the accelerating cart again experience a fictitious force in addition to the real forces.  At the bottom of the loop the fictitious centrifugal force has magnitude mass´(25m/s)²/10m = mass´62.5m/s² = mass´6.4g in the downward direction.  This force is added to the rider's weight to yield an apparent weight of mass´7.4 g.

On the top of the loop the some of the carts kinetic energy has been converted into potential energy.
(1/2)mv_top² = (1/2)mv_bottom² - mgh.
v_top² = v_bottom²-2gh = (25m/s)² - 2´9.8m/s²´20m = 233(m/s)².
v = 15.3 m/s.
The fictitious centripetal force the rider experience is now directed upward and has magnitude m v²/r = mass´(15.3 m/s)²/10m = mass´23.4m/s² = mass´2.4g.  Subtracting the real weight of the rider, which points in the downward direction, we obtain an apparent weight = mass´1.4g in the upward direction.  Things seem to "fall" upward, towards the bottom of the inverted car.

Halfway up the loop the weight points straight down, but the fictitious centrifugal force points horizontally toward the outside of the loop.  There is also a tangential fictitious force because the cart is slowing down.  The magnitude of this force is mg and it is pointing up, canceling the weight. The apparent weight is therefore equal to the centrifugal force, pointing outward.

Links:

Locally, near the surface of the earth, the gravitational force is approximately constant.  In general, the gravitational force exerted by point object 1 on point object 2 is given by

.

The vector is a unit vector pointing from object 1 towards object 2, r₁₂ is the distance between the objects.  The above formula is also valid for spherical objects with r₁₂ denoting the distance between their centers.  The formula is always approximately valid as long as the distance between the objects is much larger than their respective sizes.  When the distance between the objects changes, the gravitational force does work.  The change in the gravitational potential energy is

 .

The difference in the gravitational potential energy is uniquely defined.  To agree on the value of the gravitational potential energy we need to agree on a reference point r_12i.  Often this reference point is chosen to be at infinity.  Then

U_g(r₁₂) = -Gm₁m₂/r₁₂.

The gravitational potential energy of two objects (stars, planets, satellites) separated by a distance d then is negative, U_g = -Gm₁m₂/d.  Its absolute value is equal to the amount of work that needs to be done to completely separate the objects, i.e to let d ® ¥ .

Problem:

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