HW#6a Page 1 of 6 Question 1, A 0.23 kg pinecone falls 16.0 m to the ground, where it lands with a speed of 15 m/s. (a) With what speed would the pinecone have landed if there had been no air resistance? (b) Did air resistance do positive work, negative work, or zero work on the pinecone?

Question 2:

Work down by Normal force: WN W=N*d*cos(90), angle between Normal force N (perpendicular to the

surface) and the motion (along the surface) is 90 degrees. WN=0 Work down by gravity: Wmg=mg*d*cos(90), angle between gravity (downward) and the motion (along

the surface, horizontal) is 90 degrees. Wmg=0. Work down by friction: Wf=fk*d*cos(180), angle between friction and the motion is 180 degrees.

Cos(180)=-1 , Wf = - fk*d Friction is kinetic, fk=µk*N, N=mg, W=µk*mg*d*(-1)=0.2*9*9.81*5*(-1)= -88.2 J b). Total work is equal to change of kinetic energy: Wtotal=Wfriction=Kf - Ki = -88.2 J So: Kf=Ki+W , Final Kinetic energy is equal to initial kinetic Energy plus the work down to it. The work is negative. Kf is less than Ki ½ mvf

2=½ mvi2+Wfriction , Simplify it by multiply 2/m to both side of the equation, you get:

vf2= vi

2+Wfriction*(2/m) vf2= 72+(-88.2)*(2/9) vf=5.42 m/s

c). From start till become rest, Wtotal=Krest - Kinitial=0 - Ki = - ½ m vi

2 = - ½ *9*72= -220 J This is the short way, only worries about the initial and final kinetic energy. Since fk=µk*N is a constant force, the acceleration is constant, a=fk/m= - µk*N/m From the acceleration and initial and final velocity, you can find out the total distance from the beginning till it stops, using our old friend, vf

2= vi2+2a(∆x), (Note that a is negative here.) . Wtotal=Wfriction = fk*∆x*(-1) = -220 J This is the longer way.

When you know initial and final velocity, the total work equals to the change of kinetic energy is the best way for you to find total work. (no need to worry about acceleration at all). This also works for non-straight motion, for non constant acceleration, loop, complicated track, etc. It is always true that: Wtotal=Kfinal - Kinitial See the energy flow chart from lecture notes of Chapter 7 &8 part 2.

HW#6a Page 2 of 6 Question 3 ,

Don’t call all angles θ. Pay attention which ones are the angles between force and motion, and which one is the incline angle. In this question I call the incline angle θ . θ=60 degrees

Work down by Normal force: W=N*d*cos90 = 0 Work down by gravity: Wmg=mg*d*cos(90-θ), the angle between gravity (downward) and

the motion (along the incline) is 90-θ=30 degrees. Wmg=mg*d*cos(30)=5*9.81*4*cos(30) = 170 J. Also work done by gravity is always equal to mg times the motion distance along the gravity force direction: (hi-hf). Wmg= mg*(hi-hf)= mg*(∆h) |∆h|= d*sinθ =d*sin60= d*cos30 = 3.46 m, It lowered height by 3.46m and gravity did positive work. KE increased and PE decreased. Wmg= mg*(hi-hf)= 5*9.81*3.46 = 170 J

Work down by friction: Wf=fk*d*cos180, Wf=fk*d*(-1),

Friction is kinetic, fk=µk*N, N=mg*cosθ =mgcos60 Wf=µk*mg* cos(60)*d*(-1)=0.4*5*9.81*cos(60)*4*(-1)= - 39.2 J. Wtotal=Wmg+Wfriction= 170-37.2 =130.8 J Wtotal =Kf - Ki Kf = Ki +Wtotal = 0 + 170 - 39.2 =130.8 J ½ mvf

2 = Kf so, vf2 =2 Kf /m vf

=sqrt(2 Kf /m)=sqrt (2*130.8/5) = 7.23 m/s Attention: when we add total work, since work and energy are scalar, it doesn’t matter which direction the force is. Just add positive and negative numbers together. Positive total work increases the kinetic energy. And negative total work reduces kinetic energy. c) Since the acceleration is constant. It is not too hard to find the time, the average velocity is easy to deduce. vave = ½ (vi+vf) =1/2 vf , would be half the final velocity, or 3.615 m/s, so the time would be 4.00 m/3.615 m/s = 1.11 s. However, if the slope of the incline was not constant, the acceleration would not be constant, and there is no obvious way to figure out the average velocity and the time. So the energy way is very convenient for tracks with various shapes.

HW#6a Page 3 of 6 Question 4. At an amusement park, a swimmer uses a water slide to enter the main pool. If the swimmer starts at rest, slides without friction, and descends through a vertical height of 2.61 m, what is her speed

at the bottom of the slide? 7.16 m/s Total work done by non-conservative force is zero.

Don’t always assume: vf =sqrt (2gh), this is only true when vi=0 and when there is no work done by friction and external work. Some Numbers used in the solution methods are different from numbers printed in this question part. Learn the methods and practice the methods yourself using the red numbers to get the answers in the green boxes. Question 5,

Total work done by non-conservative force is zero.

HW#6a Page 4 of 6 Question 6,

More insights: Energy is scalar but not vector. Energy and work have no direction. Even if the initial velocity was not upward, Initial Kinetic energy Ki is still = ½ mvi

2, no matter which direction vi is. Final Kinetic energy Kf is still = ½ mvf

2, no matter which direction vf is. If the only force which did work is the gravity force, like in this question. The total work is the work done by gravity: Wtotal = mg*(hi-hf). If h decreased, gravity did some positive work to increase the kinetic energy. It doesn’t matter that gravity force is in vertical direction and the initial velocity may have some horizontal component. Simply: Wtotal =½ mvf

2 -½ mvi2 If you are only asked about the total final speed (velocity size), there is NO need

to worry about directions and x or y components. Only for calculating time, final velocity direction and distance in x or y directions, you need to follow what you did in chapter 4. Since we learned potential energy U already, we also don’t need to think too much about the work done by mg any more. The work done by gravity is represented by the loss of the potential energy. Wmg = U i - Uf If no friction, nor other force does work (like, man, motor…) Simply : Ei = Ef U i + Ki = Uf + Kf Or U i - Uf = Kf - Ki If non-conservative forces (friction, motor, human…) did some work Wnc, (some HW 6b) Simply : Wnc = Ef - Ei Ef = Ei+Wnc

HW#6a Page 5 of 6 Question 7,

Again, don’t always assume: vi =sqrt (2gh), this is only true when vf=0 and when there is no work done by friction and external work. It is not true for the next question: Question 8,

More insight: Normal force doesn’t do work in this case, because it is always perpendicular to the motion direction. The angle between force and motions is 90. cos(90)=0.

HW#6a Page 6 of 6 Also in this question the skier didn’t use his mussel to do any some work. Why? Because it is an ideal frictionless situation. In such a situation, the skier can only act force perpendicular to the track, but can not apply any force to along the motion’s direction. This man is not an energy source in such a frictionless place. He moves as if he is a crate without energy power in absolute frictionless surface. That’s why it was safe to write Ef =Ei, in step 1. Wnc = 0, no friction nor external force did work.

Also, attention: when you see circular motion, or part of a circle or arc, you should think about centripetal acceleration immediately. The total net force along radius direction pointing to the center is not zero. At the bottom of the track, analyze the total force along the radius direction, pointing toward the center, and make that equals to mv2/r. That’s how you do the 3rd step. Once we find out v2 in the second step, we can happily solve the 3rd step. Question 9,

c), Similar to the Normal force in previous questions, the tension doesn’t do work in this case, because it is always perpendicular to the motion direction. Tension is in the radius direction and motion is in tangent direction. The angle between force and motions is 90 degrees. cos(90)=0. This is why we can safely write Ef =Ei There is no Work done by non-conservative force. Neither friction or external force. a), Ef =Ei , U i + Ki = Uf + Kf Lowest point, set U i = 0 , Ki =½ mvi

2 Maximum height, Kf =0 , U f = mgh

So, mgh =½ mvi2 All initial kinetic energy are converted into potential energy.

h =½ * vi2 /g =½ * (2.2)2 /9.81 = 0.247 m

b), vi = half of previous vi =1.1 m/s, h =½ vi

2 /g =½ * (1.1)2 /9.81 = 0.0617 m Notice that if the new vi is 1/2 of previous vi , the new vi

2 is only 1/4 of previous vi2

The new initial kinetic energy is ¼ of previous initial kinetic energy. the new maximum height h will be ¼ of previous h, so the new h will be 0.247/4=0.0617 m Because the initial kinetic energy is proportional to vi

2, change of v will be “squared” and contribute to the change of kinetic energy. This is also why if you double the speed, the distance for stopping the car will become 4 times.