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Announcements CAPA Set #9 due Friday at 10 pm This week in Section: Lab #3 – Momentum Reading – Chapter 7 on Momentum.

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Presentation on theme: "Announcements CAPA Set #9 due Friday at 10 pm This week in Section: Lab #3 – Momentum Reading – Chapter 7 on Momentum."— Presentation transcript:

1 Announcements CAPA Set #9 due Friday at 10 pm This week in Section: Lab #3 – Momentum Reading – Chapter 7 on Momentum

2 Two paths lead to the top of a big hill. Path #1 is steep and direct and Path #2 is twice as long but less steep. Both are rough paths and you push a box up each. How much more potential energy is gained if you take the longer path? A) noneB) twice as much C) four times as muchD) half as much hh d 2d2d #1#2 θ φ Clicker QuestionRoom Frequency BA

3 Consider the same two paths as in the previous problem. Is the amount of work done the same along each path? A) Yes B) No, more work is done on the steeper path. C) No, more work is done on the longer path. D) Probably not, but the details are needed to know which is larger. h h d 2d2d #1 #2 θ φ Clicker QuestionRoom Frequency BA

4 Force = -kx Hooke’s “Law” k = “spring constant” units = N/m x = displacement of spring Large k = stiff spring Small k = floppy spring Springs Spring in “relaxed” position exerts no force Compressed spring pushes back Force Stretched spring pulls back Force

5 PE = ½ kx 2 Stored Energy in Springs Compressed spring has elastic potential energy Stretched spring also has elastic potential energy If no change in any other energy, then  PE = W ext =F spring x d However F spring = -kx changes as we stretch the spring Use average F spring = -½ kx Elastic potential energy:

6 F = -kx, PE = ½kx 2 A force of magnitude F stretches a spring through a displacement x. The force is then increased so that displacement is doubled. What is the magnitude of force needed to double the displacement and what will be the effect on the potential energy of doubling displacement? A)Force doubles, PE doubles. B)Force doubles, PE quadruples. C) Force quadruples, PE doubles. D) Force doubles, PE halves. E) Force quadruples, PE quarters. Clicker QuestionRoom Frequency BA

7 Which of the following types of potential energy can be negative? A)Gravitational Potential Energy B)Elastic Potential Energy C)Both of the above D)Neither of the above Clicker QuestionRoom Frequency BA PE gravity = mgy

8 A block of mass m is released from rest at height H on a frictionless ramp. It strikes a spring with spring constant k at the end of the ramp. How far will the spring compress (i.e. x)? KE i + PE i + W frict + W external = KE f + PE f 00 0 + mgH + 0 + 0 = 0 + (0 + ½ kx 2 ) gravitational elastic

9 Elastic Potential Energy – Does Not Have to Look Like a Spring

10

11 What is Power?

12 The average power can be written in terms of the force and the average velocity: Power is a rate of energy per time Units are Joules / second = Watts

13 Raise a box of mass 2 kg at constant velocity by a height of 6 meters over a time interval of 2 seconds. 6m 2 kg How much energy is required to do the work? W ext =  PE = mgh since  KE=0 W ext = 2 kg x 9.8 m/s x 6 m = 117 Joules How much power is required to do the work in that amount of time? Power = W ext / time = 117 J / 2 s = 59 Watts Also W ext = F ext x d  F ext = 117J/6m = 19.6 N Also Power = F ext v = 19.6 N x (6m/2s) = 59 Watts

14 The average household in the USA uses 2.7 billion Joules of electrical energy each month. Approximately how many 100 Watt lightbulbs would need to be left on all the time to amount to that energy consumption? xx A)1 lightbulb B)10 lightbulbs C)100 lightbulbs D)1000 lightbulbs E)10,000 lightbulbs Clicker QuestionRoom Frequency BA A kiloWatt-hour is a unit of energy (Power x time). 1 kW-hour = 3.6 x 10 6 Joules Energy = Power x time = (10 x 100 W) x (30x24x60x60 s) ≈ 2.6 x 10 9 Joules

15 m1m1 m2m2 v 1i v 2i Before the collision: After the collision: m1m1 v 1f m2m2 v 2f v 1f = final velocity of mass 1 v 2f = final velocity of mass 2 Collisions These collisions should obey Newton’s Laws

16 Total kinetic energy remains constant during the collision KE is conserved or ΔKE = 0 No energy converted to thermal or potential energy Elastic Collision

17 Macroscopic objects never have perfectly elastic collisions but it’s often a good approximation. Inelastic Collision: KE is not constant; some is converted to another kind of energy; typically heat or deformation. Totally Inelastic Collision: KE is not conserved, and the objects stick together after collision. Demonstration

18 Kind of like kinetic energy (associated with motion). However, not in units of Joules (instead kg m/s) and so not an energy. In addition to total energy being conserved in every collision, the total momentum vector is also conserved! Conservation of Momentum (in one-dimension) New Conserved Quantity  Momentum


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