CONSERVATIVE AND NON-CONSERVATIVE FORCES The work done by a conservative force depends only on the initial and final position of the object acted upon. An example of a conservative force is gravity. The work done equals the change in potential energy and depends only on the initial and final positions above the ground and NOT on the path taken.

Friction is a non-conservative force and the work done in moving an object against a non-conservative force depends on the path. For example, the work done in sliding a box of books against friction from one end of a room to the other depends on the path taken.

For mechanical systems involving conservative forces, the total mechanical energy equals the sum of the kinetic and potential energies of the objects that make up the system and is always conserved.

What is the impact Velocity? dropped from rest V?

Conservation of Mechanical Energy The mechanical energy (Emec) of a system is When no external forces act on a system WNC = 0 Conservation of Mechanical Energy

If there is no friction, the speed of a roller coaster will depend only on its height compared to its starting height. y

See Energy Skate Park applet

Solve with Conservation of Energy or Projectile motion Theory Solve with Conservation of Energy or Projectile motion Theory? (assume WNC = 0)

In real life applications, some of the mechanical energy is lost due to friction. The work due to non-conservative forces is given by: Eo = Ef +WNC

Gravity/Kinetic Energy m d vo = 0 V = ? q h h = d sin(q)

W = Fd W = ΔEk (joules) (newton) (meter) WORK = FORCE X DISPLACEMENT http://ngsir.netfirms.com/englishhtm/Work.htm W = Fd W = ΔEk

M Assume d = 5.13m What is the Work done? Calculate Vf?

Incline plane W = ΔEk + ΔEp W = Fd Assume: d = 5.3 m h2 = 1.6 m What is Vf? W = ΔEk + ΔEp W = Fd

A At point A: UA + KA At point B: KB UA + KA= KB B = 42.9 m/s The tallest and fastest roller coaster in the world is the Steel Dragon in Japan. The ride includes a vertical drop of 93.5 m. The coaster has a speed of 3 m/s at the top of the drop. Neglect friction and find the speed of the riders at the bottom in km/h? A At point A: UA + KA At point B: KB UA + KA= KB vA = 3 m/s hA = 93.5 m hB = 0 m B = 42.9 m/s = 155 km/h !!!

A 20-kg sled rests at the top of a 30˚ slope 80 m in length. If μk= 0.2, what is the velocity at the bottom of the incline? m = 20 kg θ = 30° x = 80 m μk= 0.2 U0 = Kf + WFf

PE0 = mgh0 = 20(9.8)(40) = 7840 J Ff = μkFN WFf = Ff x m = 20 kg θ = 30° x = 80 m μk= 0.2 h = x sin θ h = 80 sin 30° = 40 m PE0 = mgh0 = 20(9.8)(40) = 7840 J x h Ff = μkFN = μkFgy = μkFgcos30° = (0.2)(20)(9.8)cos30° = 34 N WFf = Ff x = 34(80) = 2720 J U0 = Kf + WFf KEf = PE0 - WFf = 7840 - 2720 = 5120 J = 22.6 m/s

WNC = Ef - E0 = KB - (PEA + KA) = - 4416.5 J b. Find the work done by non-conservative forces on a 55 kg rider during the descent if the actual velocity at the bottom is 41 m/s. vA = 3 m/s vB = 41 m/s hA = 93.5 m hB = 0 m m = 55 kg WNC = Ef - E0 = KB - (PEA + KA) = - 4416.5 J