Physics 121. Tuesday, February 19, Physics 121. Tuesday, February 19, Physics 121. Course announcements. Topics:

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1 Physics 121. Tuesday, ebruary 19, avy Lt. Ron Candiloro's /A-18 Hornet creates a shock wave as he breaks the sound barrier July 7. The shock wave is visible as a large cloud o condensation ormed by the cooling o the air. A smaller shock wave can be seen orming on top o the canopy. It is possible or a skilled pilot to work the plane's throttle to move the shock wave orward or at. /A-18 Hornet breaking the sound barrier. Photo by John Gay Physics 121. Tuesday, ebruary 19, Topics: Course announcements Quiz ork and Energy: Deinition ork done by a Variable orce Kinetic Energy Physics 121. Course announcements. On Thursday ebruary 28 between 8 am and 9.30 am the irst midterm exam o Physics 121 will be held. The material covered on the exam is the material covered in Chapters 1-6 o our text book. The location o the exam is Hubbell auditorium. There will be a normal lecture ater the exam (at 9.40 am in Hoyt). A ew remarks about the exam: You will be provided with an equation sheet with all important equations used in Chapter 1-6. There will be no numerical questions on the exams. All problems must be solved in terms o the variables provided. o calculators required. Practice exams are provided on the EB. 1

2 Physics 121. Course announcements. Homework set # 4 is due on ebruary 23 at 8.30 am. There will be no homework set due on Saturday March 1. Homework set # 5 will be available on the EB on Thursday morning, ebruary 28, at 8.30 am. The most eective way to work on the assignments is to tackle 1 or 2 problems a day. I you run into problems, please contact me and I will try to help you solve your problems (Physics 121 problems that is). Physics 121. Quiz lecture 9. The quiz today will have 3 questions. ork and energy. In the next ew weeks, we will not discuss any new physics, but develop tools to simpliy how we use our understanding o the orce laws and the laws o motion to understand and/or predict the outcome o experiments. e will start with deining the concept o work and the concept o energy. 2

3 ork and energy. hen a orce is applied to an object, it may produce a displacement d. The work done by the orce is deined as =! i! d = d cos! where is the angle between the orce and the displacement d. ork. Consider the deinition o : =! i! d = d cos! The work done by the orce is zero i: (a) v0 v1 d = 0 m (no displacement) = 90 (orce perpendicular to the displacement). (b) v0 v1 ork: positive, zero, or negative. ork done by a orce can be positive, zero, or negative, depending on the angle : I 0 < 90 (scalar product between and d > 0) the speed o the object will increase. (a) v0 v1 = 90 (scalar product between and d = 0) the speed o the object will not change. (b) v0 v1 I 90 < 180 (scalar product between and d < 0) the speed o the object will decrease. 3

4 ork: units. The unit o work is the Joule (abbreviated J). Per deinition, 1 J = 1 m = 1 kg m 2 /s 2. There are many important examples o orces that do not do any work. or example, the gravitational orce between the earth and the moon does not do any work! ote: in this case, the speed o the moon does not change. Power. In many cases, the work done by a tool is less important than the the rate with which the work can be done. or example, explosive devices get their properties rom being able to do a lot o work over a very short period in time. The same amount o work done over a longer period o time might not lead to destruction. Power: units. Power o deined as work per unit time: P = d dt The unit o power is the att, abbreviated by a. Per deinition: 1 = 1 J/s = 1 kg m 2 /s 3 The power you consume at home is oten expressed in terms o kh, which is the use o 1 k o power or 1 hour. 4

5 ork. An example problem. A block o mass M is drawn at constant speed a distance d along a horizontal loor by a rope exerting a orce at angle above the horizontal. Compute (a) the work done by the rope on the block, and (b) the coeicient o kinetic riction between block and loor. ork. An example problem: step 1. The requirement that the object moved with constant speed tells us that the net orce acting on it must be 0. Thus the net orces in the x and y directions:! x = cos" # k! y = + sin" # Mg.. must be zero. ork. An example problem: step 2. The normal orce can be determined based on the act that the net orce in the vertical direction must be zero: = Mg! sin" Based on the now know normal orce we can determine the rictional orce (kinetic riction since the block is moving): ( ) k = µ k = µ k Mg! sin" 5

6 ork. An example problem: step 3. The riction orce is also directly related to the applied orce by considering net orce in the horizontal direction, which has to be zero. This requires that k = cos! The work done by the riction orce is negative (since direction and displacement are in opposite direction): =! k i! d =! k d =!d cos" ork. An example problem: step 4. The work done by the applied orce is equal to =! i! d = d cos! The net work done (due to the applied orce and the riction orce) is equal to zero. total =! i = + = 0 i This is not really a surprise. since the net orce on the object is equal to zero. ork done by a varying orce. In most realistic cases, we need to consider the work done when the orce is varying (both in magnitude and direction) as unction o time and/or position. In this case, we can still use the same approach as we just discussed by breaking up the motion into small intervals such that the path is linear and the orce is constant during the intervals considered. 6

7 ork done by a varying orce. ork. A inal remark. Do less work by thinking beore starting! Consider the work done by all orces acting on the pendulum when it moves rom position 1 to position 2. r y-axis r - h 2 During this motion, the angle between the path and the net h orce changes. hat am I to do? x-axis = i! d! = (! +! g )i d! = i! d! +! g i d! x 1! g i d! =! g h =!mgh! i d! = x = h( 2r! h) The work-energy theorem. e have already seen that there is a connection between the work done by a orce and the change in the speed o the object: I > 0 J: speed increases I = 0 J: speed remains constant I < 0 J: speed decreases 7

8 The work-energy theorem. Consider the bus starting rom rest (v 1 = 0 m/s) and having an acceleration a = net /m. The velocity at a later time t will be equal to This relation can be used to determine the time t at which the bus reaches a certain velocity v: t = v/a. v( t) = v 0 + at = at The work-energy theorem. The displacement at this time t is equal to! d = x t = v $ " # a% & = x 0 + v 0 t at 2 = 1 2 a v 2! $ " # a% & = 1 v 2 2 a The work done by the orce during this period is equal to Kinetic =! net i!! 1 v d = ( ma) 2 $ " # 2 a % & = 1 2 mv2 Energy K The work-energy theorem. e conclude: The net work done on an object is equal to the change in its kinetic energy. In the case o the bus: net d = 0.5mv mv 1 2 8

9 The work-energy theorem. An application. An object with mass m is at rest at time t = 0 s. It alls under the inluence o gravity through a distance h. hat is its velocity at that point? Solution: ork done by the gravitational orce = mgh. Change in kinetic energy = 0.5mv 12. ork-energy theorem: mgh = 0.5mv 1 2 or v 1 = (2gh) y-axis y0 = 0 m v0 = 0 m/s y1 = -h v1 =? That s all! Thursday: conservation o energy. Unusual Spherules on Mars Credit: Mars Exploration Rover Mission, JPL, USGS, ASA 9