Michael's Physics Page
Sunday, December 2, 2012
Monday, November 26, 2012
Monday, November 12, 2012
Monday, November 5, 2012
Thursday, November 1, 2012
Lab Assessment F=ma
Analysis
part 1.
Acceleration
(m/s2)
|
Force
Applied (N)
|
0.4
|
0.4
|
0.8
|
0.8
|
1.2
|
1.2
|
1.6
|
1.6
|
2.2
|
1.9
|
The linear relationship tells us that the larger
the net force is, the larger the acceleration will be.
Questions
1) The slope of the line should indicate the mass of
the cart.
a.
Explain
why the slope is the mass. Show also that the units of the slope correlate with
the units of the mass.
The slope represents the mass because of the
format of the equation. The original equation for the Net Force is: Fapp
- Ffriction =
mass*acceleration. Solving for Fapp yields
to: Fapp =
mass*acceleration + Ffriction. This is in slope-intercept
form, y=mx+b where ‘y’ represents Fapp, ‘m’ represents mass, ‘x’
represents acceleration and ‘b’ represents Ffriction. You can verify
that the slope represents the mass by considering the units of the y-axis and
the x-axis. Y(N)/X(m/s2) Ã
Y(kg*m/s2)/X(m/s2) Ã
kgm/s2 * s2/m = kg
b.
What
is the slope? What is the actual mass? Find the experimental error.
Slope = .8525kg
Actual Mass =
807g = .807kg
Experimental
Error = 5.3% error
c.
We
don't expect more than 15% error. Identify at least two solid reasons for the
error.
·
Wheels
not balanced
·
Friction
between string and wheel
·
Force
of friction along table wasn’t constant
2) The y intercept is the force of friction. Explain
why! what is the average force of friction as seen on your graph.
The y-intercept
is the force of friction because it always exists, even when there is no force
pulling the cart. Therefore at 0N of force, friction still takes place. The
average force of friction was approximately 0.123.
3) The friction force you found in 5 is only the
average force. Use the smallest force (least acceleration) and largest force
(highest acceleration) to compute from an equation the friction force in each
case. Which one seems to be larger? Do you expect this result.
Smallest force (least acceleration) à .4 = .8525*.4 + b à b = 0.059N
Highest force (most acceleration) à 1.9 = .8525*2.2 + b à b = 0.0245N
The smallest force yields a larger force of acceleration. I
expected this to happen because the force of friction tends to have more of an
impact on an object with less applied force. Possibly, the more force you apply
on an object the less of an impact the force of friction will have.
Analysis for
part 2
Mass
(g)
|
Acceleration
(m/s2)
|
807.2
|
1.8
|
1307.2
|
1.3
|
1807.2
|
0.9
|
2307.2
|
0.7
|
2807.2
|
0.6
|
3307.2
|
0.5
|
1) What kind of relationship do we expect?
Inverse relationship
2) Does this relationship agree with Newton's second
law which is
Forcenet = ma or a=Fnet/m
Forcenet = ma or a=Fnet/m
YES! The equation
we got was y=929.51/x.926
In this
equation, ‘y’ represents acceleration, 929.51 represents the Fnet
and ‘x’ represents the mass.
Conclusion
With the help of toy cars, weights,
sensors and computers we were able to model the type of work Isaac Newton did
to derive the famous equation F=ma.
To show the relationship between
Force and Acceleration, we increased the applied force on a toy car and using
sensors determined its acceleration. Performing five consecutive tests with
increasing forces we concluded that there is a direct positive relationship
between the Fapp and the acceleration. Plotting (acceleration, Fapp)
as coordinates we produced a best-fit line that gave us the direct relationship
between the ‘y’ (force) and ‘x’ (acceleration). The equation was in the form of
slope-intercept, as expected, because the equation represented Fapp
= mass*acceleration + Ffriction. Comparing the slope-intercept form
of an equation with that of Fapp we concluded that the slope of the
line represented the mass. We verified this using the slope formula (change in
y / change in x), which yielded N/(m/s2) = kg. The setup of our experiment was prone to many
errors in calculations, such as: unbalanced wheels on carts, change of friction
on surface, friction between string and wheel (that the weights were attached
to). In future experiments we should strive
to eliminate as many errors as possible so our results can be more accurate.
We were also able to show that Mass
and Acceleration have an inverse square relationship, as Newton stated with his
formula a=F/m. We performed the same tests as in the first experiment; however
this time the applied force remained constant and we altered the mass on the
toy cart. Using the same sensors we determined the acceleration while using 5
different masses, and the graph produced was indeed that of an inverse square
relationship. The form of the equation we received was y=k/x, where ‘y’
represents the acceleration, ‘k’ represents the constant force and ‘x’
represents the mass.
Wednesday, September 19, 2012
Physics Classroom Kinematics Practice
Physics Classroom Kinematics Practice
Michael Matias
September 19th, 2012
1.
An airplane accelerates down a runway at 3.20 m/s2 for 32.8 s until is finally lifts off the ground.
Determine the distance traveled before takeoff.
d= v_i*t+1/2*a*t^2→d=0+1.6*(32.8)^2→d= 1721m
2.
A car starts from rest and accelerates uniformly over a time
of 5.21 seconds for a distance of 110 m. Determine the acceleration of the car.
d= v_i*t+1/2*a*t^2→110=0+1/2*a*(5.21)^2→110=13.57a→a=8.1m/s^2
3.
Upton Chuck is riding the Giant Drop at Great America. If
Upton free falls for 2.6 seconds, what will be his final velocity and how far
will he fall?
v_f=v_i+a*t→v_f=0-9.8*2.6→v_f=-25.48m/s
d=(v_i+v_f)/2*t→d=(-25.48)/2*2.6→d=-33.1m
4. A race car accelerates uniformly from 18.5 m/s to
46.1 m/s in 2.47 seconds. Determine the acceleration of the car and the
distance traveled.
v_f=v_i+a*t→46.1=18.5+a*2.47→27.6=2.47a→11.17m/s^2
d=(v_i+v_f)/2*t→d=(18.5+46.1)/2*2.47→d=79.79m
5. A feather is dropped on the moon from a height of
1.40 meters. The acceleration of gravity on the moon is 1.67 m/s2. Determine the time for the feather to fall to
the surface of the moon.
d= v_i*t+1/2*a*t^2→1.4=0+.8*t^2→t^2=1.75→t=1.32s
6. Rocket-powered sleds are used to test the human
response to acceleration. If a rocket-powered sled is accelerated to a speed of
444 m/s in 1.8 seconds, then what is the acceleration and what is the distance
that the sled travels?
v_f=v_i+a*t→444=1.8*a→a=246.67m/s^2
d=(v_i+v_f)/2*t→d=444/2*1.8→d=399.6m
7. A bike accelerates uniformly from rest to a speed
of 7.10 m/s over a distance of 35.4 m. Determine the acceleration of the bike.
d=(v_i+v_f)/2*t→35.4=3.55t→t=9.97s
v_f=v_i+a*t→7.71=0+9.97*a→a=.77m/s^2
8. An engineer is designing the runway for an
airport. Of the planes that will use the airport, the lowest acceleration rate
is likely to be 3 m/s2. The takeoff speed
for this plane will be 65 m/s. Assuming this minimum acceleration, what is the
minimum allowed length for the runway?
v_f^2=v_i^2+2ad→〖65〗^2=0+2*3*d→4225=6d→d=704.2m
9. A car traveling at 22.4 m/s skids to a stop in
2.55 s. Determine the skidding distance of the car (assume uniform
acceleration).
d=(v_i+v_f)/2*t→d=22.4/2*2.55→d=28.56m
10. A kangaroo is capable of jumping to a height of
2.62 m. Determine the takeoff speed of the kangaroo.
v_f^2=v_i^2+2ad→0=v_i^2+2*-9.8*2.62→v_i^2=51.352→v_i=7.2m/s
11. If Michael Jordan has a vertical leap of 1.29 m,
then what is his takeoff speed and his hang time (total time to move upwards to
the peak and then return to the ground)?
v_f^2=v_i^2+2ad→0=v_i^2+2*-9.8*1.29→v_i^2=25.284→v_i=5.03m/s
Time to peak -> d=(v_i+v_f)/2*t→1.29=5.03/2*t→t=.51s
Hang Time =
2*.51 = 1.02s
12. A bullet leaves a rifle with a muzzle velocity of
521 m/s. While accelerating through the barrel of the rifle, the bullet moves a
distance of 0.840 m. Determine the acceleration of the bullet (assume a uniform
acceleration).
v_f^2=v_i^2+2ad→〖521〗^2=0+2*a*0.84→a=161572m/s^2
13. A baseball is popped straight up into the air and
has a hang-time of 6.25 s. Determine the height to which the ball rises before
it reaches its peak. (Hint: the time to rise to the peak is one-half the total
hang-time.)
v_f=v_i+a*t→0=v_i-9.8*3.125→v_i=30.625m/s
d=(v_i+v_f)/2*t→d=30.625/2*3.125→d=47.9m
14. The observation deck of tall skyscraper 370 m
above the street. Determine the time required for a penny to free fall from the
deck to the street below.
d= v_i*t+1/2*a*t^2→370=0+1/2*9.8*t^2→370=4.8t^2→t=8.78s
15. A bullet is moving at a speed of 367 m/s when it
embeds into a lump of moist clay. The bullet penetrates for a distance of
0.0621 m. Determine the acceleration of the bullet while moving into the clay.
(Assume a uniform acceleration.)
v_f^2=v_i^2+2ad→0=〖367〗^2+2*a*.0621→0=134689+0.1242a→a=-1084452m/s^2
16. A stone is dropped into a deep well and is heard
to hit the water 3.41 s after being dropped. Determine the depth of the well.
d= v_i*t+1/2*a*t^2→d=0+1/2*-9.8*〖3.41〗^2→d=-57m
17. It was once recorded that a Jaguar left skid
marks that were 290 m in length. Assuming that the Jaguar skidded to a stop
with a constant acceleration of -3.90 m/s2, determine the speed of the Jaguar before it
began to skid.
v_f^2=v_i^2+2ad→0=v_i^2+2*-3.9*290→v_i^2=2262→v_i^2=47.6m/s
18. A plane has a takeoff speed of 88.3 m/s and
requires 1365 m to reach that speed. Determine the acceleration of the plane
and the time required to reach this speed.
v_f^2=v_i^2+2ad→〖88.3〗^2=0+2*a*1365→a=2.86m/s^2
d=(v_i+v_f)/2*t→1365=88.3/2*t→t=30.92s
19. A dragster accelerates to a speed of 112 m/s over
a distance of 398 m. Determine the acceleration (assume uniform) of the
dragster.
v_f^2=v_i^2+2ad→〖112〗^2=0+2*a*398→a=15.76m/s^2
20. With what speed in miles/hr (1 m/s = 2.23 mi/hr)
must an object be thrown to reach a height of 91.5 m (equivalent to one
football field)? Assume negligible air resistance.
v_f^2=v_i^2+2ad→0=v_i^2+2*-9.8*91.5→v_i^2=42.4m/s=94.6mi/hr
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