27-Sep-2014: Coefficients of friction stuff

a) Static friction with cup and water

1. set up the equirement as shown in the pictures. Add water in the cup until the one of the block starts moving. Weight the mass of the water and block. 

2. Putting one more block on the blocks before and repeat the last step. Repeat four times until the number of block is four.

 Record the data of the mass of block and mass of the water each time.

type the data in logger pro and caclculate the Normal force and Tension for each time. Use linear fit steacher the graph. The slope of the graph is the coefficient between the table and blocks. As shown below, the =0.257.

 b) kinetic friction with force sensor

Pull the force sensor and recored the data.

The slope of the graph shows the acceleration  which is 1.77m/s^2.

And the normal force of those four blocks is 5.71N. then a=(T-N)/m

 c) static friction from angle to get sliding

Set up the equirement as shown. Raise up the higher end of the track little by little until the block starts moving. Measure the angle between the track and table θ . 

Do a free body diagram.

mg*sinθ-mg*cosθμ=ma

μ=(g*sinθ-a)/g*cosθ=0.373

 d) kinetic friction from sliding block on a steep ramp

Set up the equirement as shown. Recored the data.

Make a FBD, the mass of the block is 100g, the mass of the block is 128g. The angle is 28degree. Acceleration shown in the Velocity Vs. time grap( the slope ) is 3.59ms^2
then a=(Mg-mg*sinθ-mg*cosθ*μ)/(M+m)
μ=(Mg=mg*sinθ-a(M+m))/mg*cosθ=0.373

e) predicting acceleration of a two-mass system—one

on a friction ramp, the other hanging.

Use the same formula in d) a=(Mg-mg*sinθ-mg*cosθ*μ)/(M+m) perdict the acceleration is a=2.92m/s^2

27-Sep-2014: Non-Constant acceleration problem/Activity: elephant

A 5000-kg elephant on frictionless roller skates is going 25 m/s when it gets to the bottom of a
hill and arrives on level ground. At that point a rocket mounted on the elephant’s back generates
a constant 8000 N thrust opposite the elephant’s direction of motion.
The mass of the rocket changes with time (due to burning the fuel at a rate of 20 kg/s) so that the
m(t) = 1500 kg – 20 kg/s·t.

Find how far the elephant goes before coming to rest.

Ideas: 

1. Newton’s 2nd law gives us the acceleration of the elephant + rocket system as a function of 

time: a(t)= Fnet

m(t)

= !8000 N

6500kg!20 kg

s t

= !400

325!t

m

s

2 

2. You can integrate the acceleration from 0 to t to find ∆v and then derive an equation for v(t): 

v(t)= v0 + a(t)dt.

0

t

! 

3. You can integrate the velocity from 0 to t to find ∆x and then derive an equation for x(t): 

x(t)= x0 + v(t)dt.

0

t

! 

4. You can solve v(t) to find the time at which v = 0. 

5. Then you can use the time you derived above in 4 and plug that into your expression for x(t) 

to find how far the elephant goes. 

Numerical integration: 

1. Open up a new Excel spreadsheet and enter the following:

2. Set things up so that the time increments by 0.1 seconds for at least 220 rows 

3. Input a formula into cell B2 that will let you calculate the acceleration at any time. 

4. Fill that formula down to cell B3. 

5. In cell C3 calculate the average acceleration for that first 0.1 s interval. 

6. In cell D3 calculate the change in velocity for that first time interval. (Use A3-A2 for the 

time rather than 0.1 s, so you can calculate this stuff using different time intervals if you so 

choose.) 7. In cell E3 calculate the speed at the end of that time interval.

8. In cell F3 calculate the position of the elephant. That will be 

xat the beginning of the interval +vinterval·!t. If you have done this correctly you should be able 

to “Fill Down” the contents of Row 3 to the rest of the spreadsheet and determine when and 

where the elephant comes to rest, AND the answers should agree closely with what you go 

doing things analytically. 

9. Change the time interval to 1 second instead of 0.1s and see if it makes a difference. 

10. Change the time interval to 0.05 s instead of 0.1s and see if it makes a difference

A2 enter 0
E2 enter 25
F2 enter 0
A3=A2+0.1
B2=-400/(30-A2)
C3=(B2+B3)/2
D3=400*(LN(325-A3)-LN(325-A2))
E3=E2+D3
F3=F2+E3*(A3-A2)
then fill down

find out the speed +0 to -0 betweeb 19.6s - 19.7s. And the elephant tarveld 247.4m during this time.

determination of an unknown mass (Practice the caculation of Incentainly)

set up the equipment as shown bleow

Measure both side of the angle and read the scales.

Measurement shown in the picture bleow.

θ1=44°±2°    F1=4.7N±0.5N

θ2=52°±2°    F1=5.5N±0.5N

Mass of the unknow is 

m= (cos(θ1）*F1+cos(θ2)*F2)/g

m= (cos(44°）*4.7+cos(52°)*5.5)/g

dm=|sin(θ1)F1/g|*dθ+|sin(θ2)F2/g|*dθ+|cos(θ1)/g|*dF+|cos(θ2)/g|*dF

m=0.6905kg±0.09518kg

Mwasuring the Density of Metal Cylinders (Pracite the caculation of Incertainly)

1. Measre the diameter of each cylinder. d
2. Measure the hight of each cylinder.  h
3. Measure the mass of each cylinder. m

copper
d=1.26±0.01cm
h=5.00±0.01cm
m=56.6±0.1g

aluiminum
d=1.43±0.01cm
h=4.83±0.01cm
m=21.0±0.1g

brass
d=1.59±0.01cm
h=4.8±0.01cm
m=18.1±0.1g

ρ=4m/πd^2h
dρ=I∂ρ/∂mI*dm+I∂ρ/∂dI*dd+I∂ρ/∂hI*dh
devide left side by ρ.
devide right side by 4m/πd^2h.
dρ/ρ=dm/m+2dd/d+dh/h

then the density of each metal is:
copper:            9078.53±0.02kg/m^3
aluiminum:  2707.14±0.02kg/m^3
brass:      1899.12±0.02kg/m^3

14-sep-2014: Trajectores

1.Set up the apparatus as shown.

2. launch the ball from a readily identifiable and repeatable point near the top of the inclined ramp.notice where it hits the floor.
3.Tape a piece of carbon paper to the floor around where the ball landed. launch the ball five times from the same place as before and verify that the ball lands in the virtually the same place each time.

4. Determine the height of the bottom of the ball when it launches, and how far out from the table's edge it lands.
Height h=93.7cm. How far from the edge of the table x=72.5cm.
5.Dteremine the launch speed of the ball from your measurments. sketch the dimensions clearly and show your calculation.
known g=9.8m/s^2
vertical: (1/2)gt^2= h    =>   t=0.437s
horizontal: vt=x     =>     v= 1.658m/s

6.imagine attaching an inclined board at the edge of the lab table such that now the, launched at the same spot as before, will strike the board a distance d along the board. Derive an sxpression that would allow you to determine the value of ad given that you know v0 and the angle.
7.Place a board such that it touches the end of the lab table and the floor. Put a heavy mass on the floor at the foot of the board and use duct tape to fix the mass in place. attach a piese of carbon paper to your board such that it " surrounds" the spot where you expect your ball to land. Make appropriate measurements to determine the angle of board. Then run the experiment, launching your ball five times from the same spot.

the angle between the board and horizontal is measured as α =48 degree.

tanα=H/X

H=(1/2)gt^2

X=v0t

=> t= 0.3758s

=>X=0.623m

cosα=X/d

=> d=0.93m

8.Determine the experimental value of your landing distances d and report your experimental value as d +_ σ
d+_0.03m
9.Compare your experimental and theoretical values for d. Comment on soures of uncetainty or error in the experiment.

14-sep-2014: Modeling the fall of an object falling with air resistance

1.Determing the relationship between air resistance force and speed.
We have an expectation taht air resistance force aon a particular object depends on the object's speed: F=kv^n.
Dorp the coffee filters from tthis balcony.
If you drop coffee filters at H. Reachs terminal speed and you measure vertical velocity, your actual know the velocity of resistance force at the speed.

falling down with one filter

falling down with two filters

falling down with three filters

falling down with four filters

falling down with five filters

Data analysis

Enter the solpe of each time 

graphing with power fit

k=0.00435

n=2.3466

2.Modeling the fall of an object including air resistance.

We can use Excel to model the alll of an object with air resistance.

we start with the folling conditions.

Cell A2 enter 0, A3 enter =A2+0.1, fill down till the value shows 5.

Cell B2 enter 9.8, B3 enter 9.8-(0.0043566*(D3^2.347)/0.0092)), then fill down.

Cell C3 enter =B2*A3.

Cell D3 enter =D2+C3.

Since the measurement is not accurate, the K and N value is not the exactly, the table filling down with errors occured.

6-Sep-2014:Free Fall Lab- determination of g

1. Set up the euqrientment.

2.Record the date.

3. Date analysis.

Open an excel file. 

In cell A1 enter Time. In cell A2 enter 0.In cell A3 enter =A2=1/60. if you highlight cell A3 and several cells below that, then choose the Edit/ Fill/ Down menu EWxcel automatically fill the cells below.

Choose a bot to be your t=0s dot.

In cell B1 enter Distance. In cell B2 enter0. In cell B3 enter the position of 2nd dot from your spark tape. Enter the rest of the dot positions in cell B4,B5,ect.

In cell C1 enter delta x. In cell C2 enter=(B3-B2). Fill down this to the cells below this on.

In cell D1 enter Mid- interval time. In cell D2 enter =A2=1/120. fill down

In cell E1 enter Mid interval speed. In cell E2 enter =(C2/(1/60)).

Dart we got form our spark tape.

4. Graphing Data

Select your data in columns D and E. Click on the Chart Wizard on the top toolbar to start making the graph. Choose an SY(scatter) graph with the points NOT connected.Give appropriate titles (with units) for your graph and graph axes.place the chart as a new sheet.

under the chart menu sellect Add Trenline, choose Linear. Under the Pitions tab you click "OK" this should give you a line and equation on the graph.

Double click on the equtaion. Use the Nuberes formatting to get a reasonable number of decimal places in your answers. Use the Font menu to make the equtaion large enough that it will show up nicely when you print your graph.

Select Colums A and B. follow the same setps above Except when you get to Add Trendline choose a polymonial fit of order 2.

Linear fit for Mid-interval speed

Find out the g=9.296m/s^2

Plymonial fit for Distancce VS. time

We have unexplained spread in our data. All of our values are too low. systematic error.