Conservation of Energy

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? • Suppose I lift straight up at constant speed. • Suppose slide up a 22o slope at constant speed.

5. Lift Example 1 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/s2) 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(q) = (2kg)(9.8m/s2)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/s2 for 0.25 s, then lift at constant speed, then slow it back down at a rate of 2.1 m/s2 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. SF=ma F-mg=ma, so F= mg+ma = (2kg)(9.8m/s2)+(2kg)(2.1m/s2) = 23.8N d=½ at2 =½ (2.1m/s2) (.25)2 = 0.065625m W1=(23.8N)(0.065625m)= 1.561875J Part 3 – slowing down SF=ma F-mg=ma, so F= mg+ma = (2kg)(9.8m/s2)+(2kg)(-2.1m/s2) = 15.4N d again= 0.065625m W3=(15.4N)(0.065625m)= 1.010625 J Part 2 – constant speed SF=ma F-mg=0, so F= mg = (2kg)(9.8m/s2) = 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=Fxdx+Fydy+Fzdz • using components of the vectors, • or W=Fdcos(f) • where F and d are magnitudes, and f = 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 (r), 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 (r), 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