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course Phy 201
q001. On a graph of velocity v (in cm/sec) vs. clock time t (in sec):What are the velocity and clock time corresponding to the point (4, 12)?
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Velocity= 4cm/sec time=12 sec
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@& On a graph of v vs. t, the first coordinate is that of the independent variable t, the second coordinate being the dependent variable v.
So for this point t = 4 sec and v = 12 cm / sec.
If you have reversed this order, as many students do, it won't be a serious problem on this question. However do make note of the correct order.
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What are the velocity and clock time corresponding to the point (9, 32)?
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Velocity= 9cm/sec time=32sec
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If these points correspond to the velocity of a ball rolling down an incline, describe as fully as you can what you think happens between the first event (corresponding to the first point of the graph) and the second event (corresponding to the second point of the graph).
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As the ball travels down the incline the velocity steadily increases. This means that the speed increases.
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What is the change in velocity between these two events?
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9cm/sec-4cm/sec= 5cm/sec
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What is the change in clock time between these two events?
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32sec-12sec= 20sec
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What is the average velocity for the interval between these two events?
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(9cm/sec+4cm/sec)/2= 6.5cm/sec
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What is the average rate at which the velocity changes, with respect to clock time, between these two events?
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(9cm/sec-4cm/sec)/(32sec-12sec)=(5cm/sec)/10sec= 0.5cm/sec^2
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What is the displacement of the object between these two events?
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5cm
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@& vAve = (change in position) / (change in clock time), so
(change in position) = vAve * (change in clock time
You found a correct average velocity, and you know the time interval, so what do you conclude is the change in position?
*@`q002. A 5 kg mass accelerates at 2 m/s^2. What is the net force acting on the object?
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5kg*2m/s^2= 10kgm/s^2
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`q003. A net force of 5000 kg m/s^2 acts on a 100 kg mass. What is the acceleration of the mass?
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5000kg m/s^2= 100kg * a so divide both sides by 100kg and a= 5m/s^2
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`q004. A force of 400 Newtons is exerted on an automobile as it is pushed through a distance of 100 meters. How much work is done on the automobile?
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Dw=400newtons*100meters dw=40,000newton meters
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`q005. A certain pendulum requires .5 Newtons of force for every centimeter it is pulled back (recall pulling back the pendulum hanging from the tree limb, using the rubber band).
How much force would be required to pull the pendulum back 5 cm, 10 cm and 15 cm?
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0.5Newtons/cm*5cm= 2.5Newtons
0.5Newtons/cm*10cm=5Newtons
0.5Newtons/cm*15=7.5Newtons
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What is the average force required between pullbacks of 5 cm and 15 cm?
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(5Newtons + 2.5Newtons)/2= 3.75Newtons
@& Right idea, but at 15 cm the force is 7.5 N so the average force turns out to be 5 N.*@
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How much work is done between these two positions?
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3.75Newtons*10cm= 37.5Newton cm
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`q006. A system rotates through 12 rotations in 4 seconds, first coming to rest at the end of this interval.
How quickly is it rotating, on the average? (The answer is as obvious as it should seem, but also be sure to interpret this as a rate of change with respect to clock time, and carefully apply the definition of average rate)
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12rotations/4seconds= 3rotations/second
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Is it speeding up or slowing down?
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Slowing down
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At what rate is it doing so? (Again attempt to interpret as an average rate of change of an appropriate quantity with respect to clock time).
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(0 rotations-12rotations)/4seconds= -3rotations/second
@& Frequent answers are 12 rot./ 4 sec. = 3 rot./ sec or -3 rot / sec.
However that's the average rate at which it's rotating, not the average rate at which is it slowing down.
By how much does its rate of rotation change? You know that its average rate of rotation is 3 rot / sec and its final rate is 0, so what are its initial and final rates of rotation?
How much slowing down does it therefore do on this interval, and how long does it take to do so?
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`q007. At the initial point of an interval a 7 kg mass is moving at 5 meters / second. By the end of the interval it has gained an additional 200 kg m^2 / s^2 of kinetic energy.
How much kinetic energy does it therefore have at the end of the interval?
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KE= (1/2)*(7kg)*(5m/s)^2= 87.5kgm^2/ s^2
87.5kgm^2/s^2+ 200kgm^2/s^2= 287.5kgm^2/s^2
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Slightly more challenging question: How fast is the mass therefore moving at the end of the interval?
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287.5kg m^2/s^2= (1/2)*(7kg)(v)^2
82.14m^2/s^2=v^2
V= 9.06m/s
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`q008. An object begins an interval with a kinetic energy of 20 000 kg m^2 / s^2, and ends the interval with a kinetic energy of 15 000 kg m^2 / s^2.
By how much did its kinetic energy change on this interval? (The answer is as obvious as it might seem, but be careful about whether the answer is positive or negative).
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15,000kg m^2/s^2 - 20,000kg m^2/s^2= -5000kg m^2/s^2
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More challenging: During this interval, how much work was done on the object by the net force?
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-5000kg m^2/s^2
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Also more challenging: If the average force on the object during this interval had magnitude 200 Newtons, then what was its displacement during this interval?
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-5000Newtons= 200Newtons*ds
Ds= -25cm
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`q009. A mass of 200 grams hangs from one side of a pulley, and another mass from the other side. The gravitational force pulling down on this mass is about 200 000 gram cm / s^2, and the tension in the string pulling it upward is about 180 000 gram cm / s^2.
Pick either upward or downward as the positive direction.
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upward
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Using + for your positive direction and - for your negative direction, what is the gravitational force on this object?
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-200,000gram cm/s^2
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Using + for your positive direction and - for your negative direction, what is the tension force on this object?
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180,000gram cm/s^2
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Using + for your positive direction and - for your negative direction, what is the net force on this object?
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-200,000gram cm/s^2+180,000gram cm/s^2= -20,000gram cm/s^2
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Using + for your positive direction and - for your negative direction, what is the acceleration of this object?
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-20,000gram cm/s^2= 200gm*a
a= -100cm/s^2
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If the object's displacement during a certain interval is +30 cm, then according to your choice of positive direction, is the displacement upward or downward?
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upward
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When you multiply the displacement by the gravitational force, what is your result? Be sure to indicate whether the result is positive or negative.
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30cm*-200,000gram cm/s^2= -6,000,000gram cm^2/s^2
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When you multiply the displacement by the tension force, what is your result? Be sure to indicate whether the result is positive or negative.
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30cm*180,000gram cm/s^2= 5,400,000 gm cm^2/s^2
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When you multiply the displacement by the net force, what is your result? Be sure to indicate whether the result is positive or negative.
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-20,000gm cm/s^2*30cm= -600,000gm cm^2/s^2
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Does gravity do positive or negative work on this object?
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negative
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Does the tension force do positive or negative work on this object?
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positive
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Does the net force do positive or negative work on this object?
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netative
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Does the object speed up or slow down?
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Slow down
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How would your answers have changed if you had chosen the opposite direction as positive?
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The work done by gravity would be positive and the net force would be positive. Tension force would be negative.
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`q010. A pendulum of length 2 meters and mass 3 kg, pulled back a distance | x | from its equilibrium position, experiences a restoring force of magnitude k | x |, where k = 15 kg / s^2 * | x |. [Note that for convenience in calculation we are making some approximations here; the actual value of k for this pendulum would actually be closer to 14.7 kg / s^2, and this is so only for values of | x | which are a good bit smaller than the length. These are details we'll worry about later.]
How much force does the pendulum experience when x = .1 meter?
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0.1meter*15kg/s^2= 1.5kg meter/sec^2
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How much force does the pendulum experience when x is .05 meter, .1 meter, .15 meter and .2 meter?
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If x is 0.05 then force is 0.75kg m/s^2
0.1 then force is 1.5kg m/s^2
0.15 then force is 2.25kg m/s^2
0.2then force is 3kg m/s^2
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What do you think is the average force between | x | = .05 meter and x = .2 meter?
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( 0.75kg m/s^2+3kg m/s^2)/2= 1.875kg m/s^2
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How much work do you think would be done by this force between | x | = .05 meter and | x | = .2 meter?
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1.875kg m/s^2* 0.15m= 0.28kg m^2/s^2
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How fast would the pendulum have to be going in order for its kinetic energy to equal the result you just obtained for the work?
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0.28kg m^2/s^2= (1/2)*(3kg)*(v^2)
0.187m^2/s^2=v^2 v=0.43m/s
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If the pendulum moves from position x = .05 meter to x = .2 meter, is the direction of the force the same as, or opposite to the direction of the motion?
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Same as
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If the pendulum moves from position x = .20 meter to x = ..05 meter, is the direction of the force the same as, or opposite to the direction of the motion?
Opposite
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If the pendulum string was cut, what would be the acceleration of the 1 kg mass?
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10m/s^2
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What is the magnitude of the force exerted by gravity on the pendulum's mass? For simplicity of calculation you may use 10 m/s^2 for the acceleration of gravity.
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Force= 1kg*10m/s^2 Force=10kg m/s^2
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When x = .1 meter, what is the horizontal displacement from equilibrium as a percent of the pendulum's length?
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0.1meter
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When x = .1 meter, what is the restoring force as a percent of the pendulum's weight?
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10kg m/s^2
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What is the magnitude of the acceleration of the pendulum at the x = .15 meter point?
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10m /s^2
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`q011. The force exerted on a mass has magnitude | F | = 15 Newton / meter * x, for 0 <= x <= .25 meter.
Sketch a graph of | F | vs. x. (You might wish to start by making a table of | F | vs. x, for some appropriate values of x between 0 and .25 meter). Note the convention that a graph of y vs. x has y on the vertical axis and x on the horizontal, so that for this graph | F | will be on the vertical axis and x on the horizontal.
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Verify that the points (.06 meter, .9 Newton) and (.16 meter, 2.4 Newtons) lie on your graph.
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yes
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What is the rise between these points?
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0.16-0.06= 0.10meter
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What is the run between these points?
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2.4Newton-0.9Newton= 1.5 Newton
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@& The independent variable is position and it is the first coordinate. The dependent variable is force and it is the second coordinate.
The second coordinate is represented by the vertical axis, and the rise is the change in the vertical coordinate.
So you have the rise and run reversed.*@
What is the average slope between these points?
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0.06meter/Newton
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What is the average 'graph altitude' of the 'graph trapezoid' formed by these points?
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17.25 Newton
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What is the width of the trapezoid?
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0.24meters
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What therefore is the area of the trapezoid?
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4.14 Newton meters
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What are the graph points corresponding to x = .05 meter and to x = .20 meter?
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(0.05, 7.5) (0.20, 3)
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What is the area of the 'graph trapezoid' defined by these points?
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Average force with respect to distance
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@& You would actually multiply the average 'graph altitude' of the trapezoid by its width*@
What is the meaning of the altitude of this trapezoid?
Average force
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What is the meaning of the width of this trapezoid?
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Meters moved
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What therefore is the meaning of the area of this trapezoid?
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Force in respect to distance
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@& These quantities are multiplied, so you get average force * displacement.
What is average force * displacement?*@
`q012. For the rotation data you took in class:
What was the average rate of rotation in the trial where the added masses were at the end of the rotating beam?
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0.25rotations/second
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What would then have been the initial rate of rotation (at the instant your finger lost contact with the system)?
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1rotation/second
@& If the rate of rotation changes at a constant rate, then
What is the average rate of rotation?
Since the system comes to rest at the end of the interval, what do you conclude must have been the initial rate?
Do your initial and final rates of rotation average out to the average rate you originally found?
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Assuming that those masses were 14 cm from the center of rotation, how fast were they moving, in cm / second, at that initial instant?
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@& During one complete rotation this mass would travel one time around a circle of radius 14 meters.
The distance it travels, per revolution, is therefore equal to the circumference of the circle.
At the initial rate of rotation how fast were these masses therefore moving?
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Assuming that the masses were each 60 grams, what was the kinetic energy of each mass?
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KE= ½(60g)(0.25rotations/second) KE=15g rotation/second
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For University Physics Students only:
`q013. For the v vs. t trapezoid whose width is `dt and whose altitudes are v0 and vf:
What is the slope of the graph and what does the slope mean? Be sure to explain the entire interpretation of the slope.
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What is the area of the graph and what does the area mean? Be sure to explain the entire interpretation of the area.
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In terms of v0, vf, `dt, a and `ds what two equations do we get from the expressions for the slope and the area?
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What equation do we get when we eliminate vf from these two equations? Verify that you know how to do the algebra of this elimination.
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What equation do we get when we eliminate `dt from these two equations? Verify that you know how to do the algebra of this elimination.
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`q014. The following allow us to define work (F * `ds), kinetic energy (1/2 m v^2), impulse (F `dt) and momentum (m v):
If we solve F_net = m a for a and plug the result into the second of the equations obtained above, then solve this equation for F `dt, what is the result? Show the algebra of your solution, or verify that you can do the algebra easily.
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If we solve F_net = m a for a and plug the result into the second of the equations obtained above, then solve this equation for F `ds, what is the result? Show the algebra of your solution, or verify that you can do the algebra easily.
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@& You did very well on the majority of these questions.
See my notes and let me know if there's anything you aren't sure of. We will of course be going over this in class on Monday.*@