Centripetal Force Practice

Projectile Motion

Practice Newton Forces

Homework due November 7th, 2012

Homework due November 5, 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: F_app - F_friction = mass*acceleration. Solving for F_app yields to: F_app = mass*acceleration + F_friction. This is in slope-intercept form, y=mx+b where ‘y’ represents F_app, ‘m’ represents mass, ‘x’ represents acceleration and ‘b’ represents F_friction. 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) à Y(kg*m/s²)/X(m/s²) à 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
Force_net = ma or a=F_net/m

YES! The equation we got was y=929.51/x^.926

In this equation, ‘y’ represents acceleration, 929.51 represents the F_net 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 F_app and the acceleration. Plotting (acceleration, F_app) 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 F_app = mass*acceleration + F_friction. Comparing the slope-intercept form of an equation with that of F_app 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.

Physics Classroom Kinematics Practice

Physics Classroom Kinematics Practice

Michael Matias

September 19^th, 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 

Test review for finding average speed and average velocity

For each of the following cases, find average speed and average velocity
1. An object moves 20mto the left, then 50m to the right then again 100m left. The trip took 20s.

answers: Average speed: 8.5m/s Average velocity:3.5m/s to the left

Average Speed = 170/20 = 8.5m/s
Average Velocity = 70/20 = 3.5m/s left

2. An object moves west 40m, back east 5m , then north 15m. Trip took 2.5 minutes.
Answers: Average speed: 0.4m/s Average velocity: 0.25m/s at 23.2⁰ off the west. )

Average Speed = 60/2.5 = 24meters/minute = 0.4m/s
Average Velocity = 38.1/2.5 = 0.25m/s NW 23.2 degrees

Physics Classroom Practice

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+.5a*27.14→.5a=4.05→a=8.1m/s^2

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?

d= v_i*t+1/2*a*t^2→d=0+.5*-9.8*6.76→d=32.68m

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→a=11.17m/s^2

d=(v_i+v_f)/2*t→d=79.78m

Kinematics Review

Kinematics equations/ review guideline

Chapter 3

1.         A tennis ball is dropped from 1.2m above the ground. It rebounds to a height of 1m.

a.        With what speed does it hit the ground?

d= v_i*t+1/2*a*t^2→1.2=0*t+1/2*9.8*t^2→4.9t^2=1.2→t^2=2.4→t=1.55s

v_f=v_i+a*t→ v_f=0+9.8*1.55=15.19m/s 

b.        With what speed does it leave the ground?

v_f^2=v_i^2+2*a*d→0= v_i^2+2*-9.8*1→v_i^2=19.6=4.4m/s

2.         A car starts from rest and gets to a store that is 400m away. The car’s velocity when it gets to the store is 10m/s.  Another car that starts in the same place, is already moving at 12m/s. Which car is going to get to the store first and by how many seconds less will it travel?

CAR 1: d=(v_i+v_f)/2*t→400=(0+10)/2*t→400=5t→t=80s

CAR 2: d=(v_i+v_f)/2*t→400=12t→33.3s

CAR 2 will get to the store 46.66s before the CAR 1

3.         Engineers are developing new types of guns that might someday be used to launch satellites as if they were bullets. One such gun can give a small object a velocity of 1m/s moving it through only 2cm.

a.        What acceleration does the gun give this object?

d=(v_i+v_f)/2*t→.02=(0+1)/2*t→t=.04s 

v_f=v_i+a*t→1=0+a*.04→1=.04a→a=25m/s 

b.        Over what time interval does the acceleration take place?

d=(v_i+v_f)/2*t→.02=(0+1)/2*t→t=.04s 

4.         A rock is thrown down into a well with an initial downward speed of 3m/s. The rock hits the bottom 1.8s later.

a.        How deep is the well?

v_f=v_i+a*t→v_f=3+9.8*1.8→v_f=20.64m/s 

d=(v_i+v_f)/2*t→d=(3+20.64)/2*1.8→d=21.276m 

b.        What is the rock's speed as it hits the bottom?

v_f=v_i+a*t→v_f=3+9.8*1.8→v_f=20.64m/s 

5.         A hot air balloon is ascending at a steady velocity of 2.2m/s. An object is dropped from it.  What is the distance from the object and the balloon after 0.5s?

d= v_i*t+1/2*a*t^2→d=2.2*.5+1/2*9.8*〖.5〗^2→d=2.325m

Motion Exercises

Motion exercises

1.    A car moving at 36km/h is uniformly accelerated at the rate of 0.5m/s.

a.    What is the speed of the car 6s after the acceleration begins?

v_f=v_i+a*t→ v_f=10.83+.5*6=13.83m/s 

b.    How far does the car move during the 6s

d=(v_i+v_f)/2*t→d= (36+39)/2*6=225m 

2.    A ball rolls down a hill accelerating at 3m/s.

a.    How far did it move in 4s if it starts from rest?
     v_f=v_i+a*t→v_f= 0+3*4=12m/s
     d=(v_i+v_f)/2*t→d=(0+12)/2*4=24m

b.    What is its speed after the 4s?

v_f=v_i+a*t→v_f= 0+3*4=12m/s

3.    Pressing the brake of a car it slows down from 30m/s and comes to stop during 6s.

a.    What is its stopping distance?

d=(v_i+v_f)/2*t→d=(30+0)/2*6=90m

b.    What is its acceleration?

 v_f=v_i+a*t→0=30+a*6→ -30=6a→a=-5m/s^2

4.    Two cars are moving at 30m/s. One car starts accelerating at 0.2m/s.(the other car continues in the same speed). What will the distance between the two cars be 4s after the acceleration started?

Car 1: v_f=v_i+a*t→v_f=30+.2*4=30.8m/s^2

            d=(v_i+v_f)/2*t→d=(30+30.8)/2*4=121.6m

Car 2: d=(v_i+v_f)/2*t→d=(30+0)/2*4=60m

Distance between the two: 121.6-60 = 51.6m

5.    An object moving at 20m/s starts accelerating at -0.5m/s covering a distance of 20 m. How long does this motion take?

d= v_i*t+1/2*a*t^2→20=20*t+1/2*-0.5t^2→0=-0.25t^2+20t-20→t=1.01s