Conservative and Non Conservative Forces

Conservation of Energy

Energy in a closed system is always conservedA closed system means that neither mass nor

energy can be transferred to or from outside this system

It means that the total energy within the system must remain the same.

However, energy can transform from one form to another

Heat energy is taken from the drink to break the bonds in the ice

Kinetic energy of the stone is transformed into heat by the friction

Conservation of mechanical energy

Consider a body in a CLOSED LOOPA closed loop is a system where the body

returns again and again to the same position-eg: a mass on a spring

a bouncing ball uniform circular motion

Conservative Forces

If the body is under the influence of a CONSERVATIVE FORCE

.. The body will have the same KINETIC ENERGY, (and velocity) at the beginning and at the end of the loop.

Since WORK is the CHANGE in KINETIC ENERGY, the net work in a closed loop is..

.. ZeroNo net work is done in a closed loop

Principle 1

Consider the closed loop A - B - ASince the total work over the loop must be zero for

a conservative force .... We know that

W1 = -W2

A

B

W1 W2

Now consider that W1 and W2 both have the same starting and finish points.Since in the previous situation, W1=-W2, then in this situation, W1=W2

Principle 2

A

B

W1 W2The work done is the same irrespective of which path is taken.This is called: PATH INDEPENDENCE

Conservative Forces

Gravity is the best example of a conservative forceConsider a ball that it is thrown directly upwards

and allowed to fall back to the ground

Principle 1- work over a closed loop

W = Fx.cos(q)W(up )

= mgh.cos(180)= -mgh

W(down) =mgh.cos(0)= mgh

Total Work= -mgh + mgh= 0

Principle 2 – path independence

Recall that it required the same amount of work for gravity to bring the leaf down to the ground, irrespective of its path.

qh

h(cosq)

Non - conservative Forces

Friction is the best example of a non conservative force

Consider a crate that is pushed a distance ‘x’ over a ‘rough’ floor

And then pushed back to it’s original position

Principle 1- work over a closed loop

W = Fx.cos(q)W(right) = (kN)x.cos(180) = - (kN)x

W(left) = (kN)x.cos(180) = - (kN)x

Total Work = -(kN)x + -(kN)x

= -2(kN)x

Fk Fk

x

Principle 2 – path independence

B

A

x

As seen previously, if the distance between A and B ‘x’ ...

Then the work done by friction between A and B is -(kN)x

Consider now a different path between A and B ..

Since friction is always opposite to the direction of motion, the work done on this new path is ...

-(kN)x

SummaryCONSERVATIVE FORCESDo no net work over a closed loop

(kinetic energy is conserved on returning to the same position each time)

Do equal work between two points, irrespective of its path

NON CONSERVATIVE FORCESDo net work over a closed loop

(kinetic energy is lost on returning to the same position each time)

Do more work over longer distances

Problem Sheet 4:

Solutions1 A small disk of mass 4 kg moves in a circle of radius 1 m on a horizontal surface, with coefficient of kinetic

friction of .25. How much work is done by friction during the completion of one revolution? • A disc moving with friction in a circle • As we know with frictional force, the force exerted on the disc is constant throughout the journey, and has a

value of F k = μ k F n = (.25)(4kg)(9.8m/s 2) = 9.8N . At every point on the circle, this force points in the opposite direction of the velocity of the disk. Also the total distance traveled by the disc is x = 2Πr = 2Π meters. Thus the total work done is: W = Fx cosθ = (9.8N)(2Π)(cos180 o ) = - 61.6 Joules. Note that over this closed loop the total work done by friction is nonzero, proving again that friction is a nonconservative force.

2 Consider the last problem, a small disk traveling in circle. In this case, however, there is no friction and the centripetal force is provided by a string tied to the center of the circle, and the disk. Is the force provided by the string conservative?

• To decide whether or not the force is conservative, we must prove one of our two principles to be true. We know that, in the absence of other forces, the tension in the rope will remain constant, causing uniform circular motion. Thus, in one complete revolution (a closed loop) the final velocity will be the same as the initial velocity. Thus, by the Work-Energy Theorem, since there is no change in velocity, there is no net work done over the closed loop. This statement proves that the tension is, in this case, indeed a conservative force.

3 Calculus Based Problem Given that the force of a mass on a spring is given by F s = - kx , calculate the net work done by the spring over one complete oscillation: from an initial displacement of d, to -d, then back to its original displacement of d. In this way confirm the fact that the spring force is conservative.

• a) initial position of mass. b) position of mass halfway through oscillation. c) final position of mass • To calculate the total work done during the trip, we must evaluate the integral W = F(x)dx . To since the mass

changes directions, we must actually evaluate two integrals: one from d to –d, and one from –d to d: • W = -kxdx + -kxdx = [- kx 2]d -d + [- kx 2]-d d = 0 + 0 = 0 • Thus the total work done over a complete oscillation (a closed loop) is zero, confirming that the spring force

is indeed conservative.

Potential Energy

If

AndThen

Mechanical energy is conserved underconservative forcesKinetic energy varies under workThere must be at least another form ofenergy involved in the system

This energy is called POTENTIAL ENERGYIt is usually represented by the letter ‘U’This energy increases as kinetic energy decreases,and vice versa

This implies…

U + K = EPotential Energy + Kinetic Energy = Total Energy

dU = -dKThe change in Potential Energy is equal to the

negative change in Kinetic Energy

Ui + Ki = Uf + Kf = EThe sum of Potential and Kinetic Energies is

constant

Also …

SinceAndThen

SinceThen

W = KW = F(x).dxK = F(x).dx

U = -KU = - F(x).dx = - W

m= 0.5kgh = 10m

P=49J, K=0J

P=24.5J, K=24.5J

P=0J, K=49J

Examples