Student A:
> I'm having a hard time with the homework, is there any way that you can do actual examples of
problems in class. that would be more helpful.
General help on HWs:
As you could see that we did many actual problem during the class.
for example the incline problem,
the string hanging ball roating on the horizontal plane, (we did it again during quiz yesterday).
the planet rotating around the sun problem,
recently, many more problems with changing height, velocity, with or without frcition and so on,
as you can see that during the class we have no more time left to do more problems with numbers in
detail. I mentioned in my syllabus at the very beginning that there is no way to learn physics by
following solved examples. You need to understand and be able to follow and learn the METHODS.
Understand equations and know when to use which.
For example, I solved many example problems with constant acceleration, but there are tons of other
modified problems, which needs real understanding and choice of equations, instead of direct mimicing
solved examples.
Understanding is the key. If you have a very hard time with the HW, I really would like to spend more
time with you after the class to help you. Please see me after the class to schedule a time.
To start, let's do the following first.
In chapter 7 and 8, the most important things we learned was:
1, work changes kinetic energy. It is always ture that Wtotal = Kf- Ki = 1/2 m vf^2 - 1/2 m vi^2
2, spring force and mg are special forces (conservitive forces) when they do work they transfer energy
between KE and PE. As a result, when Wnc= 0, we have a happy joint account of Kinetic Energy +
Potential Engergy. That's when there is no friction or external forces does work,
Kf+Uf = Ki +Ui
3, If forces other than spring force or mg do any work, Wnc is not equal to zero,
K+U will change by the amount of Wnc
Kf+Uf = Ki +Ui + Wnc
UNDERSTAND all of the above first.
Now, for HW 6a #1, you were told vf and vi, so that you know kinetic energy change Kf-Ki,
what causes that change? Total Work.
Wtotal = Kf- Ki Total work = W done by mg + Work done by resistance.
So: Wmg + Wf =Kf -Ki , you know Kf, Ki and can calculate Wmg, you will find Wf.
On the other hand, It is also always ture that
Kf+Uf = Ki +Ui + Wnc
Work done by mg is not Wnc, Only resistant force contriute to Wnc
because you know vf, vi, hf, hi, you know Kf+Uf = Ki +Ui and can find Wnc, which is euqual to work
done by resistance.
HW 6a #2, and #3, You were told vi, hence you know Ki.
All you need to do is to find work done by each force, and add them together as Work total.
Then, using Wtotal = Kf- Ki you can solve Kf, hence vf
HW 6a, #4 to #6, Notice that there are no resistance, friction, and no forces other than mg
Wnc=0, we happily have Kf+Uf = Ki +Ui
#7 and #8, Notice that there are no resistance, friction. Though there are normal forces or tension
NOTICE THAT because normal foce and tention on a swing is perpendicular to the motion, cos(90)=0,
hence normal foce and tension in the swing does zero work, still Wnc=0
we happily have Kf+Uf = Ki +Ui again.
For all these problems, you only need to express Ki, Kf , Ui and Uf correctly with correct v and h, you
can solve all the unknowns.
Now, for HW 6b,
#1, wave can do work right? you can not assume Wnc=0. don't assume Kf+Uf = Ki +Ui .
How to find Wnc? You can find it by comparing Kf+Uf and Ki+Ui
Kf+Uf = Ki +Ui + Wnc
#2, Same thing, Wnc is not zero, and they tell u, how many J it is.
Kf+Uf = Ki +Ui + Wnc You can solve the unknowns.
#3, is less straightforward, but still the same principle:
Kf+Uf = Ki +Ui + Wnc
notice that Uf = mg hf, and hf is related to the distance travel on the incline by sin or cos theta? we did
that in class.
Notice that work of friction = Wnc= -fk * d
Put them all in the same equation, you can solve d.
#5, still the same principle:
Kf+Uf = Ki +Ui + Wnc
just notice that U is now 1/2 k deltax^2
and Wnc= Wfk= - fk * deltax
#6, find out the KE and PE before and after the motion and ask yourself does KE+PE change?
..
see me after class.
Topic: HW 6a #1
A 0.07 kg pinecone falls 14.9 m to the ground, where it lands with a speed of 14 m/s.
(a) With what speed would the pinecone have landed if there had been no air resistance?
Dr. Man: When it says falls, it means the pinecone falls from rest. Initial velocity is
ZERO when it starts to fall.
Topic: HW6a #2b
A 8 kg crate is sliding at a constant speed of 7 m/s along a frictionless horizontal
table, when it encounters a friction patch (s=0.5, k=0.4).
a.) Find the work done by each force acting on the crate, after it has slid 5 m:
work done by normal force:J
work done by gravity: J
work done by friction: J
b.) Find the speed of the crate after sliding this distance.
c.) Some time later, the crate comes to rest. How much total work is done on the crate,
from start to finish?* (from when it enters the friction patch to when it stops)
Student A: For 2b we are looking for the Vf and i did Vf=SQRT2ax my a=9.8m/s^2
x=5m and when i did this i got 9.899 and rounded it to 9.9. my roommate
had the exact same problem and got the exact same answer and got it
right and mine is wrong
(This is nearly identical to a problem you did on HW#5. Now we will do it with the
work method.)
Dr. Man: My answer to you:
First, I never taught an euqation such as:
Vf=SQRT(2ax) in my class.
Actually, during the class on last Thur, I again emphasized not to memorize special results
for special cases, such as v0= 0 etc. The equation I taught was Vf^2=Vi^2 +2a(delta x )
You can't simply make Vi=0
and ASSUME that Vf^2 =+2a(delta x ).
you also should notice that delta_x is not always = x. In this case, right, delta_x = 5m.
but what is a? positive or negative?????"
What is Vi????
More importantly, as we emphasized during last week, both on Tue and Thur., now
that we know the energy method, which is easy to use in many cases, can you solve
this problem using the energy method?
Total work done on the object is = how many Joules?
positive or negative?
Don't forget that change of kinetic energy = total work.
1/2 m Vf^2 - 1/2 m Vi^2 = Work (total)
One more thing:
how could the a=9.8m/s^2 ?
This is on horizontal table. Back to the very beginning:
Identify the object.
Identify all the forces on it: mg, Normal force, friction...
What causes acceleration?
Accleleration in which direction, positive or negative?
Or
Identify the object.
Identify all the forces on it: mg, Normal force, friction...
Find the work done by each forces.
Find the total work (positive or negative?)
Then, don't forget the change of kinetic energy = total work.
A lot more understanding is needed. Come to my office hours for more help
Topic: HW6a, #3
A 5 kg crate is placed at rest on an incline plane (s=1, k=0.6), whose angle is 60o. It then
slides 5 m down the incline.
a.) Find the work done by each force on the crate after sliding this distance:
work done by normal force:
work done by gravity:
work done by friction force:
b.) Using your results from part a, find the speed of the crate after sliding this distance.
c.) Using the work approach only, could you find the time it takes to slide this distance?
Discuss the limitations of using work vs. using forces to solve problems.
Student A: For part A) I got the first two, but I'm having trouble finding the Ff.
Isn't it equal to uk*N? With N being equal the y-component of gravity?
Dr Man: It asks you to find the work done by friction force:, not the Friction force itself.
You have no problem calculating fk after midterm1.
Don't forget work = force * distance * cos(angle between this force and motion).
For kinetic frction, this angle between frction and motion is 180 degree and Work of friction= ???
Topic: HW 6a #8c
At a local playground, a child on a swing has a speed of 2.2 m/s when the swing is
at its lowest point.
a.) To what maximum vertical height (above the bottom of the swing) does the child rise,
assuming he sits still and "coasts"?
b. Reanswer part a, with the initial speed of the child being half as much.
c.) During this process, how much work is done by the tension in the rope?
(What assumptions did you make in part a?)
Student A: Find the tension in the rope during this process. I am having a hard time
with this part. I'm not sure what it means when it says "what are the assumptions made
from part A"....When I did part A the mass canceled out, so the total energy for the
systems was 3.92J.
Dr. Man: Pay attention that 8c asks you to find the WORK done by the tension
during this process. First, Work = force * distance * angle between force and motion.
Work done by tension is = Tension * distance * angle between tension and motion velocity.
What is the angle between tension and motion velocity. Notice that Tension is always
in the radius deirection and velocity is always in the Tangential direction and the angle
between them is always ??? (recall the angle between normal force and motion in a
circlular track). Second, when you calculate 8a and 8b, what did you assume Wnc
to be? If you used Ei=Ef for 8a and 8b, you noticed that you already assumed Wnc =0.
(all work done by non concervertive force is =0) Is that true and why?
Topic: HW 6b #1
A block of mass m slides from rest on a frictionless loop-the-loop track, as shown in Figure 8-32.
(a) What is the minimum release height, h, required for the block to maintain contact with the
track at all times? Give your answer in terms of the radius of the loop, r.
For this question, please read the chapter 8 lecture notes.
All answers are solved there.
for the syntax error, as we mentioned many times in previous homework,
you need to write an expression of your answer using the known variables and numbers.
In webassign introduction, when the two sides of a right angle triangle has length a and b respectively, they ask
you the length of the hypotenuse c. you know c =sqrt(a square + b square).
Then in the box to answer what is c, you should write sqrt(a^2 + b^2)
or
(a^2+b^2)^(0.5)
for HW 2b, the location of the person who jumped up will vary as a fuction of time as
y= 5.0+3*t-0.5*g*t^2, for example,
you just put 5.0+3*t-0.5*g*t^2
in the box
For recent HW 5a, you solve velocity of the box on the table. mV^2/r =T and Tention T =Mg for the box under
the table attached with a sting, where the m is the mass of the box on the table, and M is the mass under the
table. so, mV^2/r =Mg, v=sqrt(Mgr/m), you should put
sqrt(Mgr/m) in the box.
For this question, if you find h=&%$ r, you just put
&%$ r in the box.
Actually, the first practice in the webassign intruduction should already prepared students to answer thiskind of
question.
Topic: HW 6b #4
A 28 kg block slides with an initial speed of 15 m/s up a ramp inclined at an angle of 35o
with the horizontal. The coefficient of kinetic friction between the block and the ramp
is 0.3. Use energy conservation to find the distance along the incline that the block slides
before coming to rest.
Student A: Using: Ki + Ui + Wnc = Kf + Uf
I'm trying to find work done by friction where
W_f = (Fk)(d)
W_f = (0.3)(N) cos theta (d)
If I don't have d I can't find work done by friction, then I can't find h_f
I've defined these variables:
friction coeffecient = 0.3
V_initial = 15
V_final = 0
hi = 0
hf = what we want to find.
Came to a road block, not sure what to do about finding Work done by friction.
Dr. Man:
The distance along the incline it climbs is d. Then what is the height? how is d related to hf?
sin or cos? We did that today during the class.
hf = d cos (theta) ? hf = d sin (theta) ?
For the work done by friction:
first is it negative or positive????
W_f = (Fk)(d)*cos(180)!!!!! not cos(35)
The angle between fk and motion is always 180 degree.
Hence cos(180)=-1
W_f = - (Fk)(d) We again reviewed this today!
(5 times already) :)
Now, Wf, hf and everything else are related to d, you can solve d by plugging then into equation:
Ki + Ui + Wnc = Kf + Uf
Student B: But if you plug in the values for hf (d*sin theta) and Wf (fk*d cos180) then
you'll have a "d" on both sides of the equation and they cancel out, no? I'm obviously
not getting how to solve for d...
Dr. Man: No, When you have d in both sides of equation, for example:
8d - 9= 7-5d
How can d cancel out? Notice that some terms have d and some terms do not have d.
And you definitely can NOT divide d on both sides to cancel them.
Again basic algebra here:
To solve an equation like this,
you need to move all items with d to one side and to move all items without d to the other side.
8d - 9= 7-5d
hence 8d +5d= 7+ 9
hence 13d =16
d =16/13
This is why we emphasize math readiness so much from the very beginning of this course.
To practice more is helpful. Pay close attention to positive or negative signs when you move
items and solve equations.
Topic: HW 6b #5
A 1.80 kg block slides on a rough, horizontal surface. The block hits a spring
with a speed of 2.00 m/s and compresses it a distance of 14.0 cm before coming to rest.
If the coefficient of kinetic friction between the block and the surface is µk = 0.560, what is
the force constant of the spring?
Student A: Ok, I know I'm supposed to plug in the potential energy for a spring
(1/2k*delta_x^2) somewhere but am I solving for "k"? And how can I do that if I don't
know the distance in order to solve for the Work of friction? I cant solve for two unknowns...
Dr. Man: good question. Draw a figure for yourself, identify the initial and final position
for this part of motion that they are asking about. When it contacts the spring, that is
your initial position, you know v0 there and delta_x was zero at the contact. when it
compresses the spring for 14.0 cm, that is your final position. You know vf there is zero,
because it comes to rest. and you know delta_x (the amount of spring compressing
distance is 14.0 cm). Now compare the initial and final position, what is the distance
of moving for the object? Also 14.0 cm, right? (it didn't move back and force before the
speed reduce to zero for the first time). As a result, during this distance, what is the work
done by friction? don't forget to convert 14cm to meter.
Student B: I'm still unsure about this problem, solving for work is not what the
question is asking, and my value for k is incorrect. Can someone please elaborate
on how to do this problem?
Dr. Man: in case there is no any work done by friction or any external forces.
Wnc=0
You have Kf +Uf.= Ki+Ui
It is very clearly written on lecture notes that:
K = 1/2 m v^2 Kinetic Energy
U = 1/2 k deltax^2 Potential energy stored in compressed spring
Notice when K =0 and when U = 0.
Since it is horizontal, mgh is the same and doesn't need to be considered.
In that case, you can solve k the spring constant, from Kf +Uf.= Ki+Ui , since you know,
Kf, Ki and you know deltax...
Now, since you do have friction, Wnc is not zero.
Kf +Uf.= Ki+Ui +Wnc
You need to include the negative work done by friction,
which is what? - fk *distance
And i just answered that distance is the same as deltaX here.
Try to be able to follow all the equations and calculations during lectures is very important