Work -   Work is calculated by multiplying the force applied by the distance the object moves while the force is being applied.   W = Fs

The unit of work is the Newton•meter, also called the joule (J) The unit of work is the Newton•meter, also called the joule (J). The equation we will use is: W = (F cosϴ)s

Ex. 1 - Find the work done by a 45 Ex. 1 - Find the work done by a 45.0 N force in pulling a luggage carrier at an angle ϴ = 50.0° for a distance s = 75.0 m.

Ex. 2 - A weight lifter is bench pressing a 710 N barbell Ex. 2 - A weight lifter is bench pressing a 710 N barbell. He raises the barbell 0.65 m above his chest and then lowers it the same distance. What work is done on the barbell during the lifting and lowering phase.

Ex. 4 - A flatbed truck accelerating at a = +1 Ex. 4 - A flatbed truck accelerating at a = +1.5 m/s2 is carrying a 120 kg crate. The crate does not slip as the truck moves s = 65 m. What is the total work done on the crate by all the forces acting on it?

Work done to an object results in a change in the kinetic energy of the object. This relationship is called the work-energy theorem.

W = Fs = mas vf2 = v02 +2as can be solved for as W = Fs = mas   vf2 = v02 +2as can be solved for as. as = 1/2 (vf2 - v02), this second term is substituted in the first equation. Fs = 1/2 m vf2 - 1/2 m v02

Fs = 1/2 m vf2 - 1/2 m v02 Work equals final kinetic energy minus initial kinetic energy.   KE = 1/2 mv2   The unit is the joule.

The Work-Energy Theorem. W = KEf - KE0 = 1/2 mvf2 - 1/2 mv02

Ex. 5 - A space probe of mass m = 5 Ex. 5 - A space probe of mass m = 5.00 x 104 kg is traveling at a speed of v0 = 1.10 x 104 m/s through deep space. The engine exerts a constant external force of F = 4.00 x 105 N, directed parallel to the displacement. The engine fires continually during the displacement of s = 2.50 x 106 m. Determine the final speed of the probe.

Ex. 6 - A 54 kg skier is coasting down a 25° slope Ex. 6 - A 54 kg skier is coasting down a 25° slope. A kinetic frictional force of fk = 70 N opposes her motion. Her initial speed is v0 = 3.6 m/s. Ignoring air resistance, determine the speed vf at a displacement 57 m downhill.

The work-energy theorem deals with the work done by the net external force, not an individual force (unless it’s the only one).

The gravitational force is a force that can do positive or negative work.   W = (mg cos ϴ°)(h0 - hf) = mg(h0 - hf)

Ex. 8 - A gymnast leaves a trampoline at a height of 1 Ex. 8 - A gymnast leaves a trampoline at a height of 1.20 m and reaches a height of 4.80 m before falling back down. Determine (a) the initial speed v0 with which the gymnast leaves the trampoline and (b) the speed of the gymnast after falling back to a height of 3.5 m.

Gravitational Potential Energy is energy due to the distance an object is able to fall.   PE = mgh PE is also measured in joules.

Wgravity = mgh0 - mghf = PE0 - PEf

The work done by the gravitational force on an object does not depend on the path taken by the object. This makes gravitational force a conservative force.

•A force is conservative when the work it does on a moving object is independent of its path. •A force is conservative when it does no net work on an object moving around a closed path, starting and finishing at the same point.

Nonconservative forces are those where the work done does depend on the path. Kinetic frictional forces and air resistance are two examples.

In the work-energy theorem both conservative and nonconservative forces act on an object.   So: W = Wc + Wnc.

If work done is equal to the change in KE and Wc is due to gravitational force, then Wnc = (KEf – KE0) + (PEf – PE0)

 

Conservation of Mechanical Energy   Wnc = (KEf - KE0) + (PEf - PE0)  becomes:   Wnc = (KEf + PEf) - (KE0 + PE0)   or: Wnc = Ef - E0

Conservation of Mechanical Energy The total mechanical energy (E = KE + PE) of an object remains constant as the object moves, provided that the net work done by external nonconservative forces is zero.

Ex. 9 - A motorcyclist drives horizontally off a cliff to leap across a canyon. When he drives off, he has a speed of 38.0 m/s. Find the speed with which the cycle strikes the ground on the other side if he is 35 m below his starting point when he strikes the ground.

Ex. 10 - A 6. 00-m rope is tied to a tree limb and used as a swing Ex. 10 - A 6.00-m rope is tied to a tree limb and used as a swing. A person starts from rest with the rope held in a horizontal orientation. Determine how fast the person is moving at the lowest point on the circular arc of the swing.

Ex. 12 - The roller coaster Magnum XL-200 includes a vertical drop of 59.4 m. Assume that the coaster has a speed of nearly zero as it crests the top of the hill. Find the speed of the riders at the bottom of the hill.