3. A body experiences a change in energy when one or more forces do work on it. A body must move under the influence of a force or forces to say work was done. A force does positive work on a body when the force and the displacement are at least partially aligned. (Maximum positive work is done when a force and a displacement are in exactly the same direction). If a force causes no displacement, it does zero work. Forces can do negative work if they are pointed opposite the direction of the displacement.

4. When the load goes up, gravity does negative work and the crane does positive work. When the load goes down, gravity does positive work and the crane does negative work. mgmg F

5. r F  F  W= F  r cosθ cos(0) = ? cos(90) = ?

6. A product is obviously a result of multiplying 2 numbers. A scalar is a quantity with NO DIRECTION. A dot product is basically a CONSTRAINT on the formula. In this case it means that F and x MUST be parallel. To ensure that they are parallel we add the cosine on the end. So basically Work is found by multiplying the Force times the displacement and the result is ENERGY, which has no direction associated with it. Kind of like Area: Area = Base x Height W = F x d

7. The VERTICAL component of the force DOES NOT cause the block to move the right. ONLY the HORIZONTAL COMPONENT of the force actually creates energy or WORK. When FORCE and DISPLACEMENT are in the SAME DIRECTION you get a POSITIVE WORK VALUE. cos(0) = ? When FORCE and DISPLACEMENT are in the OPPOSITE direction, but on the same axis, you get a NEGATIVE WORK VALUE. This doesn’t give direction, it simply means that the force and displacement oppose each other. cos(180) = ? When the FORCE and DISPLACEMENT are PERPENDICULAR, you get NO WORK. cos (90) = ?

8.  A pulley system, which has at least one pulley attached to the load, can be used to reduce the force necessary to lift a load.  Amount of work done in lifting the load is not changed.  The distance the force is applied over is increased, thus the force is reduced, since W = Fd. F m

9. SI System:  Joule (N  m) British System:  foot-pound cgs System:  erg (dyne-cm) Atomic Level:  electron-Volt (eV)

10. Practice Problem: A block of mass 2.50 kg is pushed 2.20 m along a frictionless surface with a constant 16.0 N force directed at an angle of 25.0° below the horizontal. Determine the work done by: a) The Applied Force b) The Normal Force a) The Force due to gravity b) Find the net force done on the block a)31.9 J b)0 J c)0 J d)31.9 J

11. Practice Problem: A block of mass 5.0 kg starts from rest and slides 2.50 m down an inclined plane. The coefficient of kinetic friction between the block and the plane is 0.436. Determine the work done by: a) Gravity b) Friction a) The Normal Force a)61.3 J b)-46.3 J c)0 J

12. Practice Problem: Vector A has a magnitude of 8.0 and vector B has a magnitude of 12.0. The two vectors make an angle of 40 o with each other. Find A B. 73.5 J

13. Practice Problem: A force F = (5.0 i + 3.0 j ) N acts upon a body which undergoes a displacement d = (2.0 i – j ) m. How much work is performed, and what is the angle between the vectors? W = 7 J at 56.7°

14. Practice Problem: A force F = (5.0 i + 6.0 j – 2.0 k )N acts on an object that undergoes a displacement of r = (4.0 i – 9.0 j + 3.0 k )m. How much work was done on the object by the force? (Find the magnitude and direction) W = -40 J at 118.6°

15.  For constant forces  W = F r  For variable forces, the force is only constant over an infinitesimal displacement  dW = F d r  To calculate work for a larger displacement, you have to take an integral  W =  dW =  F d r

16. The area under the curve of a graph of force vs displacement gives the work done by the force. F(x) x xaxa xbxb

17. The function here must be a function with respect to “x” or “r”. Let’s look at a POPULAR force function (F=ma). Is this function with respect to “x”? NO! You can still integrate the function, it simply needs to be modified so that it fits the model accordingly.

18. Practice Problem: A force A force acting on a particle varies with x, as shown. Calculate the work done by the force as the particle moves from x = 0 to x = 6.0 m. W = 25 J

19. Practice Problem: A force acting on a particle is F x = (4x – x 2 )N. Find the work done by the force on the particle when the particle moves along the x-axis from x= 0 to x = 2.0 m. W = 5.3 J

22. Practice Problem: A spring is hung vertically, and an object of mass 0.55 kg is attached to its lower end. Under the action of the “load” mg, the spring stretches a distance of 2.0 cm from its equilibrium position. a) Find the spring constant of the spring b) Find the Work done by the spring as it stretches through this distance k = 2.7 x10 2 N/m W = -5.4 x 10 -2 J

24. Energy is expressed in JOULES (J) 4.19 J = 1 calorie Energy can be expressed more specifically by using the term WORK(W)

25.  Kinetic energy is one form of mechanical energy. Kinetic energy is due to the motion of an object.  K = ½ m v 2 › K: Kinetic Energy in Joules. › m: mass in kg › v: speed in m/s  In vector form, K = ½ m v v

26. o An object can be subject to many forces at the same time - one force may be doing positive work on the object, another force may be doing negative work, and yet another force may be doing no work at all. o The net work, or total, work done on the object (W net or W tot ) is the scalar sum of the work done on an object by all forces acting upon the object. o W net = ΣW

27. o W net = ΔK  When net work due to all forces acting upon an object is positive, the kinetic energy of the object will increase.  When net work due to all forces acting upon an object is negative, the kinetic energy of the object will decrease.  When there is no net work acting upon an object, the kinetic energy of the object will be unchanged.  (Note this says nothing about the kinetic energy.)

28. An important relationship between kinetic energy and work can be shown with the derivation below The net work done on a particle equals the change in its kinetic energy

29. Practice Problem: A 70 kg base-runner begins to slide into second base when moving at a speed of 4.0 m/s. The coefficient of kinetic friction between his clothes and the earth is 0.70. He slides so that his speed is zero just as he reaches the base (a) How much energy is lost due to friction acting on the runner? (b) How far does he slide? a)-560 J b)1.17 m

30. Consider a mass m that moves from position 1 ( y1) to position 2 m,(y2), moving with a constant velocity. How much work does gravity do on the body as it executes the motion? Suppose the mass was thrown UPWARD. How much work does gravity do on the body as it executes the motion? In both cases, the negative sign is supplied

31. The amount of Work gravity does on a body is PATH INDEPENDANT. Force fields that act this way are CONSERVATIVE FORCES FIELDS. The total amount of work done on a body that moves around a CLOSED PATH in the field will always be ZERO with conservative forces FRICTION is a non conservative force. By NON-CONSERVATIVE we mean it DEPENDS on the PATH. If a body slides up, and then back down an incline the total work done by friction is NOT ZERO. When the direction of motion reverses, so does the force and friction will do NEGATIVE WORK in BOTH directions.

32.  Power is the rate of which work is done.  One of the things we do everyday is measure how much energy we use. The rate at which we use it determines the amount we pay to our utility company. Since WORK is energy the rate at which work is done is referred to as POWER.  No matter how fast we get up the stairs, our work is the same. When we run upstairs, power demands on our body are high. When we walk upstairs, power demands on our body are lower.

33.  P ave = W / t  P inst = dW/dt  P = F v UNITS OF POWER  Watt = J/s  ft lb / s  horsepower  550 ft lb / s  746 Watts