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course Phy 121
pProblem Number 1
If a force vector has x and y components -8.7 Newtons and 6.8 Newtons, respectively, the what are the magnitude and angle the vector?
x comp: -8.7 N y comp: 6.8 N
magnitude: (-8.7N)^2+(6.8 N)^2=c^2
c=5.43 N
angle: arctan(y/x)
arctan(6.8 N/-8.7N)
=-38.01degrees
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An angle of -38 degrees is in the fourth quadrant. A sketch of this vector would show clearly that it is in the second quadrant. So there is a discrepancy between your result and what would be obvious from a picture.
The x component is negative, so you have to add 180 degrees.
Be sure you draw a picture to represent every vector in any problem, and make sure your calculations are consistent with your picture.
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Problem Number 2
You and I are pulling on a massive, initially stationary object resting on a smooth frozen pond. You pull with a force of 5.72 pounds to the North and I pull with a force of .91 pounds to the East. The object starts to move in response to our combined force.
• At what angle with the Easterly direction does the object initially move?
• How many pounds of force does the object experience from our combined pulls?
.5.72 North (y comp) 0.91 East (x comp)
arctan(5.72/0.91)= 80.96 degrees
5.72^2+0.91^2= F^2
F=5.79 lb
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Good.
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Problem Number 3
An object of mass 10 kg is subjected to a variable force F(t) (here F(t) indicates function notation, not multiplication of F by t; F(t) is the force at clock time t) for .5 seconds. The average of this force is 17600 Newtons.
Find the change in the velocity of the object in two different ways:
• First find its acceleration and then solve the motion problem to determine the change in velocity.
• Then find the impulse of the force and use the impulse to find the change in momentum, from which you can find the change in velocity.
.Fnet= m*a
a=Fnet/m
a=1760 kg/s^2
??????? I am confused on what to do next????????? Can you guide me?
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You don't appear to have done the units calculations. The units of F_net / m are Newtons / kg. A Newton is a kg m/s^2, so the N / kg unit is the same as (kg m/s^2) / kg = m/s^2.
Having found the acceleration, and knowing the change in clock time, you should be able to use the definition of acceleration to find the change in velocity.
Now what is the definition of impulse, and how would it be applied to this situation?
What does the impulse-momentum theorem say?
What is the definition of momentum, and how could you use the impulse to find the object's change in velocity?
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Problem Number 4
An object moving to the right at 3 m/s collides with an object moving at -4 m/s (i.e., to the left). The mass of the first object is 3 kg and the mass of the second is 5 kg.
After the collision, which lasts .065 seconds, the first object is observed to have velocity -5 m/s (this negative velocity is toward the left). .
• What was the average force exerted by the second object on the first?
• What was the average force exerted by the first object on the second?
• What will be the velocity of the second object after the collision?
• How do the kinetic energy totals before collision compare with those after collision?
.
‘dv1=-4 m/s-3 m/s=-7 m/s
‘dp1= m1*’dv1
=3 kg*-7m/s=-21 kg m/s
Fave=‘dp/‘dt
=-21kg m/s /0.065 sec
= -323.08 N
Good so far. All these principles apply as well to the second part of the preceding question.
Fave=-Fave
=323.08 N
‘dp2=Fave*’dt
=323.08 N*0.065 sec
=21 N sec (kg m/s)
Change in velocity-21 kg m/s/ 5kg
=4.2 m/s
Final Velocity= vf2=v02+’dv2
=-4 m/s+4.2 m/s
=0.2 m/s
KE before: 0.5 *3 kg* (3 m/s)^2 [+] 0.5 *5 kg * (-4 m/s)^2
=53.5 J
KE after: 0.5 * 3 kg * (-5 m/s)^2 [+] 0.5 *5 kg * (4.2 m/s)^2
=81.6 J
***KE after was larger by 28.1 J
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Good.
Be sure you can answer the following question: How is all this related to the impulse-momentum theorem, and how does this help illuminate the solution to the second question in the preceding problem?
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Problem Number 5
The kinetic energy of an object is equal to the work required of the net force to accelerate it from rest to its present velocity. It is easily enough shown that this work is equal to .5 m v^2, where m is the mass and v the velocity of the object. This quantity is independent of the acceleration. We therefore say that the kinetic energy of the object is .5 m v^2.
If an object of mass 88 kg is moving at 11 m/s, what is its KE?
• If its velocity doubles, what is its KE.
• By what factor does its KE increase when its velocity doubles?
.
KE=0.5* m*v^2
KE=0.5*88kg*(11m/s)^2
KE=5324 J
KE=0.5*88kg*(11m/s*2)^2
KE=21296 J
???I am confused with the wording of what factor??????
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In algebra a term is an expression added to the rest of a larger expression. A factor is a quantity which is multiplied by an expression.
So if a quantity becomes twice as great, the factor is 2.
What is the factor in this situation?
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Problem Number 6
What are x and y the components of the velocity vector obtained when we add the two following velocity vectors:
• vector A, with x and y components 4.3 m/s and -5.5 m/s, and
• vector B whose x and y components are 2.2 m/s and -2.7 m/s?
What are the magnitude and angle of the resultant vector?
.Vector A= arctan (-5.5 m/s/ 4.3 m/s)
=-51.98 degrees
Vector B= arctan(-2.7 m/s/2.2 m/s)
. =-0.024 degrees
.???? I am a tad confused here too. Should I use the R=sqrt(Rx^2+Ry^2) equation? I know that equation just stuck on this problem?????
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You need to review addition of vectors in Introductory Problem Set 5.
You also need to understand how to add two vectors, in various ways, by sketching them on paper.
You should understand from sketching why the x component of the sum of two vectors is the sum of their x components, and why the y component the sum of two vectors is the sum of their y components.
Having found the x and y components of the sum, you then use the arcTan and the Pythagorean Theorem to find the magnitude and angle of the resultant.
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Problem Number 7
An object originally moving at a constant speed is acted upon for a specified time by a constant force of 140 Newtons. At the end of the specified time the force is removed and the object proceeds at a new constant velocity.
• If the object traveled a distance of 70 meters while under the influence of the force, and if there was no dissipation of energy, then by how much would the kinetic energy (abbreviated KE) of the object increase?
.
.Fnet*’ds=‘dKE
140 N* 7 0m= 9800
???This does not seem to be all right????
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Your result is correct, except that it lack units and hence lacks meaning.
You correctly quoted the work-kinetic energy theorem and applied it. Just be sure you include units (as I see you did in the problem below).
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Problem Number 8
An upward force of 9 Newtons is required to lift an object.
• The object is lifted through a distance of 17 meters.
How much work is required?
.work=force*displacement
work=9 Newtons*17 meters
work=153 Joules
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Good.
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Problem Number 9
What is the velocity of an object which is moving at 4 m/s and which has momentum 96 kg m/s?
p=m*v
96 kg m/s=m* 4m/s
24 kg=m
??? I assumed this was supposed to say mass? I thought it already gave velocity? HELP! ??"
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You could actually get away with just saying that the velocity is 4 m/s.
But better you should state the problem as you think I intended it, as you have done correctly.
Your solution for my intended question about the mass is correct. Good work.
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Check my notes.
I did insert some questions on one problem. If the questions don't clarify the problem, please submit an appropriate question form with your best attempt and I'll try to clarify further.
As always I'm happy to answer additional questions.
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