Topic: HW 7a #1
A new record for running the stairs of the Empire State Building was set on February 3, 2003.
The 86 flights, with a total of 1576 steps, was run in 9 minutes and 33 seconds. Consider a
scenario where a man ran up 1600 steps of the Empire State Building in 10 minutes and
59 seconds. If the height gain of each step was 0.20 m, and the man's mass was 56.0 kg,
what was his average power output during the climb? Give your answer in both watts
and horsepower.
Student A: I do not know how I am suppose to relate the record with the man, to solve for h
I just multiple .2m by the amount of steps correct since there is not enough
information to solve for the hypotenuse of each step to find the length. P=W/t and
W=(1/2mVf^2+mgh)-(0+mgh), V=.2*1600steps/t.
Dr. Man: For question 1, we know the total time to finish what he did (climb such a height),
to find power, we just need to find out the total work done by the man, right?
Power = Work/ Time
This process consumes his own chemical energy in his muscle and lifted him into such a height.
It is non-conservative work. This work increases his mechanical energy from Ei to Ef
Wnc = Ef -Ei
Ef = Kf + Uf
Ei = Ki + Ui
From the start point to the stop point, the very initial velocity and the very final velocity
to be both zero, so, Ki and Kf are both zero.
Ef -Ei=??? Wnc = ??? When you use Wnc = Ef -Ei , make sure you are looking at two
points, the very bottom one and the very top one for Ei and Ef.
We only consider the work done by the man's mussel, which increased his height location to h
That means we only need to consider the total increased Potential energy.
Once you find the total work done by this person, divided by time he spends, you find the average power.
You are done.
Also notice that since it is not constant acceleration, you will not be able to find out the
instantaneous velocity at particular moments.
You can only find v average. you should not use v average to calculate KE at one point or
one particular moment.
In this case, KE at the very initial point and the stopping point are both zero.
Topic: HW 7a Question 3:
A kayaker paddles with a power output of 41.0 W to maintain a steady speed of 1.50 m/s.
(a) Calculate the resistive force exerted by the water on the kayak.
I am not sure how to relate the resistive force of water to the power. Is it required to have mass to calculate?
Dr. Man: Recall the relation between force and power we talked today.
When the kayaker paddle with a power of 41.0W
how much work does he do per second?
During that second how far did he go?
You now know work and distance and can easily find average force exert by the kayaker. (just assume the
force is in the same direction as motion, horizontal river)...
Since the speed is steady, how is resistive force related to the his paddling force?
Topic: HW 7a Question 3:
A block of mass m and speed v collides with a spring, compressing it a distance Δx. What
is the compression of the spring if its force constant is increased by a factor of four?
Give your answer in terms of Δx.
Student A: Problem #3: k=F/x ; F=p/v ; p=sqrt(2mk): k=(4*(sgrt(2mKE))/v)/x,
how am I suppose to write this in terms of deltax?"
Dr. Man: Here is some serious misunderstanding, can you tell? This student mixed spring constant k
with KINETIC ENERGY K. p=sqrt(2m *K): Here K is KE.
Anyways, we don't need this equation for this problem. He also mixed Power with momentum p.
My answer:
F=Power/v is only for constant force, not for changing force. here, P is power. Power=Work/time,
p=sqrt(2mK) is completely something else. Here p is momentum p=mv, p=sqrt(2mK) ,
K=1/2mv^2, is Kinetic energy,
Never mix them together? :)
Power = Work/time, while Momentum =mass* velocity though they both are written as P. :(
K=1/2mv^2, is Kinetic energy,
while k is spring constant, k describes how strong a spring is.
Force by spring = - k(delta_x), Potential energy in spring is U=1/2 k (delta_x)^2
Don't gamble with equations or play games by simply plugging numbers.
Again, come back to the ground, and use one of the three golden rules.
There is no friction or other NC forces in question number 3 to do any work, so it is easy:
Ef=Ei
What is the K+U at the initial point and the final point?
Do question 12 first, if you are confused. Question 11 and 12 are old questions from last HW and
the solutions are posted.
After you understand question 12, This one is actually even easier: because there is no friction,
Ef=Ei
This one is simply asking, if you have a spring which is 4 times stronger, will it
be compressed (delta x) more or less in order to stop the same moving object.
How much more or how much less.
It is asking delta_x(new ) = _______delta_x (old)
You simply only need to put a number there.
U=1/2k(delta_x)^2
All initial KE of the object converts to U.
You can use some numbers such as U = 100 J k = 2 N/m, can you find the needed delta-x for
U =100J ( to stop the object).
Next, still U=100J in order to stop the same object, but k= 8N/m, can you find out the new
deltax needed for the new spring to store 100J?
When k(new) is 4 times stronger than k(old), would
delta(x)new be more or less than delta(x)old,
By how many times?
Also, Mathematically, for a equation: U=1/2k(delta_x)^2, if U and k are known, can you
solve the equation and get delta_x in terms of U and k?
delta_x= ......U....k?
When k is 4 times larger, the delta_x needed for same U would be how many times smaller?
delta_x (new) = What number * delta_x(old)?
Answer that number in the box
Topic: HW 7a #6
Object A has a mass m, object B has a mass 2m, and object C has a mass m/2. Rank these
objects in order of increasing kinetic energy, given that they all have the same momentum.
Indicate ties where appropriate. (Use only the symbols < or =, for example A < B=C.)
KE= 1/2mv^2
Student A: Given this equation, I am getting c<A
For instance if the momentum is the same for each p= mv, and m= 5 and v= 2, then for
a- 1/2 5 (2^2)= 10
b-1/2 (2*5) (2^2)= 20
c-1/2 (5/2) (2^2) = 5
Based on that my answer should be correct...Am I looking at something the wrong way?
Dr. Man: The answer format should be C<A<B, etc. or A+B+C
They want you to compare all three of them.
They want you to rank the KE, given that they all have the SAME momentum mv.
This student assumed all velocity to be 2m/s and mass to be 5, 2*5 or 5/2.
But, for m/2 or 2m object, what would be their v in order to get the same mv as A's mv?
Remember the example in class? For A and B to have the same momentum, if mass of C is
Half of A’s mass, the velocity of C should twice of A’s velocity.
When you consider KE, you need ½ M V SQUAR．
KE of A?
Ｖsquare!!! So, KE of C will be more or less than
Dr. Man: look at the Lecture notes, page 40. the up-right corner.
p=sqrt(2mK), when K are all the same, larger m has larger p.
can you write K= a function of p and m? When p are all the same, which one will
have larger K, the larger m object or the smaller m object?
I explained when p are the same, larger m object has less velocity.
while K=1/2 m v^2, m is not that important comparing to v, because v is squared in K,
v is more important.
When p are the same, larger m object has less velocity, hence Less or larger K?
It is always very helpful to make up some numbers
m= 1, 2 or 4kg
p= 6 kg m/s
and K will be how many J for each of them?
(velocity is how many m/s for each of them?)
Dr. Man: KE = 1/2 mv^2 This equation tells you that when v is the same, KE is proportional to m.
Larger m, larger KE. However, this question said p are the same, not v. When p are the same,
larger m means less v! so it is not obvious how to decide which one has high KE from
KE = 1/2 mv^2.
, KE= ½ (mv)^2 /m , so KE =1/2 * p^2/m
this equation can tell you when p are the same, KE is reversely proportional
to m. larger m, less or more KE?
Topic: HW7a #11
A 20 kg block slides with an initial speed of 10 m/s up a ramp inclined at an angle of 30o
with the horizontal. The coefficient of kinetic friction between the block and the ramp
is 0.7. Use energy conservation to find the distance along the incline that the block slides
before coming to rest.
Student A: I added the Initial KE with the work of the non-conservative forces of gravity and
friction to try and get the final energy. KE + Wnc = Uf
(because work initial is zero and KE final is zero). I know that the Wnc is negative
because it's working against the motion, but I keep getting the wrong answer. Can someone
tell me if that's right?
Student B: Did you relate length to height? There is a lecture note on page four of Potential
Energy lecture that has a diagram that relates length to height.
Student C: Also, if you review the homework answers from 6b, you can see that she
has done this problem