1. Conservation of Energy  The work-energy theorem says  If no external force does any work on an object (or system), then the energy of the system does not change  The total energy is the same at beginning and end (and all times between)  The energy of an isolated system is conserved

2. Work  Work is the transfer of energy to an object by applying external force.  Work is not a vector. It has a sign, but no direction in space.  The work done by a force, on an object,  is positive if the force is giving energy to the object.  depends on the size of the force and the displacement through which it is applied.  In 1-D, if force is constant:

3. Work — Units  Units of work or of energy  Since  Units of work = (units of force)*(unit of distance)  Units of work = N m  also called a Joule

4. Work—Lift Example I lift a 2-kg weight up onto a 1.1 m-tall table. How much work do I do? a)Suppose I lift straight up at constant speed. b)Suppose slide up a 22o slope at constant speed.

5. Lift Example 1 a)Given d = 1.1m, what F do I need to exert? Constant velocity  No net force, so my force to lift each is F = mg = (1kg)(9.8m/s 2 ) Work done lifting each: Note: F and d in same direction (both positive or both negative) make W positive. Total work done lifting both: 21.6 J

6. Lift Example 2 b)Given h = 1.1m, d=1.1/sin(22o) = 2.936 m what F do I need to exert? Sum of forces in x- direction (along slope) =0, so F = mgsin(  ) = (2kg)(9.8m/s 2 )sin(22o) =7.342N Work done sliding: Total work done sliding: 21.6 J

7. Lift Example 3 I can’t lift the weight at constant velocity, if it isn’t moving to begin with. Suppose I accelerate it at a rate of 2.1m/s 2 for 0.25 s, then lift at constant speed, then slow it back down at a rate of 2.1 m/s 2 for 0.25 s. No extra work is done overall in speeding it up, then slowing it back down to its original speed (here zero).

8. Lift Example 3 Part 1- speeding it up.  F=ma F-mg=ma, so F= mg+ma = (2kg)(9.8m/s 2 )+(2kg)(2.1m/s 2 ) = 23.8N d=½ at2 =½ (2.1m/s 2 ) (.25)2 = 0.065625m W1=(23.8N)(0.065625m)= 1.561875J Part 3 – slowing down  F=ma F-mg=ma, so F= mg+ma = (2kg)(9.8m/s 2 )+(2kg)(-2.1m/s 2 ) = 15.4N d again= 0.065625m W3=(15.4N)(0.065625m)= 1.010625 J Part 2 – constant speed  F=ma F-mg=0, so F= mg = (2kg)(9.8m/s 2 ) = 19.6N d = rest of the distance = 1.1m – 2(0.065625m) = 0.96875 m W2=(19.6N)(0.96875 m)= 18.9875 J Total work = 1.561875J+ 1.010625 J+ 18.9875 J = 21.6J!

9. Coincidence? No! Total work done raising weight is 21.6 J, whether you lift straight up or up a slope, at constant speed or not, … because the weights have gained 21.6 J of energy from me, independent of how I go about giving it to them. 21.6 J is how much energy it takes to make this particular change (raising) in the state of the weights, no matter what.

10. Work on a Spring W = Fd  Only works if F is constant throughout displacement,  Spring force is not constant  F=-kx

11. Work on a Spring  Calculus to the rescue: If we consider a small displacement, dx, over which the force is _________________, then the small amount of work, dW, is dW = Fdx  To get the total work,

12. Spring Potential Energy  Once again, if the force does work on the spring   Where does that energy go?  Like with gravity, it is

13. Work in 2-D and 3-D  Two ways to calculate this:  W=F x d x +F y d y +F z d z  using components of the vectors,  or W=Fdcos(  )  where F and d are magnitudes, and  = angle BETWEEN F and d  NOT angle from x-axis

14. Conservation Example A 35-g ball is placed on a compressed spring, which shoots it straight up. The spring has a spring constant of 220 N/m, and is initially compressed by 3.5 cm. Neglecting drag, how high does the ball go? Do we have an isolated system? Use: h = 0.393 m or 39 cm above where it started

15. Drag  Drag is resistance to motion of an object through a fluid  If fluid is air, sometimes called air resistance  Drag with streamline, non-viscous flow depends on:  fluid density (  ), cross-sectional area of object (A), speed of object relative to fluid (v), properties of object’s surface (C).  Cross-sectional area can be thought of as the area of the shadow the object would have, if lit from the direction of the passing fluid.

16. Drag  Depends on:  density of fluid (  ), cross-sectional area of object presented to fluid (A), relative speed of object and fluid (v), properties of the object’s surface (C).  Direction: Always opposes relative motion of fluid and object  Note: This eqn doesn’t apply to viscous or turbulent flow

17. Projectiles and drag  An object moving vertically does have the same vertical motion as an object that is moving sideways too, even if vyi is same, if drag is not negligible.  Drag force has a vertical component that depends on speed, not just vy