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course Phy 241
8/06 7:00 pm" "Question: `q001. An object of mass 10 kg is subjected to a net force of 40 Newtons as it accelerates from rest through a distance of 20 meters. Find the final velocity of the object.
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Your solution:
v0 = 0 m/s, ds = 20 m, Fnet = 40 Newtons, m = 10 kg
a = 40 Newtons/ 10 kg
a = 4 m/s^2
vf = +/- sqrt(2 * 4 m/s^2 * 20 m)
vf = +/- sqrt(160 m^2/s^2)
vf = +/- 12.7 m/s
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Question: `q002. Find the value of the quantity 1/2 m v^2 at the beginning of the 20 meter displacement, the value of the same quantity at the end of this displacement, and the change in the quantity 1/2 m v^2 for this displacement.
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Your solution:
Initial value = ½(10 kg)(0 m/s)^2
Initial value = 0
Final value = 1/2 (10 kg)(12.7 m/s)^2
Final value = 806.45 kg m^2/s^2 = J
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Question: `q003. Find the value of the quantity Fnet * `ds for the present example, and express this quantity in units of kg, meters and seconds.
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Your solution:
Fnet = 40 Newtons, ds = 20 m
40 Newtons * 20 m
= 800 kg m^2/s^2 = J
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Question: `q004. How does the quantity Fnet * `ds and the change in (1/2 m v^2) compare?
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Your solution:
The quantities from both Fnet * ds and 1/2mv^2 equations is exactly identical (800 J).
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Question: `q005. Suppose that all the quantities given in the previous problem are the same except that the initial velocity is 9 meters / second. Again calculate the final velocity, the change in (1/2 m v^2) and Fnet * `ds.
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Your solution:
vf = +/- `sqrt( v0^2 + 2a ds)
+/- `sqrt( (9 m/s)^2 + 2 * 4 m/s^2 * 20 meters)
= +/-`sqrt( 81 m^2 / s^2 + 160 m^2 / s^2) =
= +/-`sqrt( 241 m^2 / s^2) =
= +/-15.5 m/s
½(10 kg)(9 m/s)^2
= 420 kg m^2/s^2= J
½(10 kg)(15.5 m/s)^2
= 1220 kg m^2/s^2 = J
1220 J - 420 J
= 800 J
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Question: `q006. The quantity Fnet * `ds and the change in the quantity 1/2 m v^2 were the same in the preceding example. This might be just a coincidence of the numbers chosen, but if so we probably wouldn't be making is bigger deal about it.
In any case if the numbers were just chosen at random and we obtained this sort of equality, we would be tempted to conjecture that the quantities were indeed always equal.
Answer the following: How could we determine if this conjecture is correct?
Hint: Let Fnet, m and `ds stand for any net force, mass and displacement and let v0 stand for any initial velocity. In terms of these symbols obtain the expression for v0 and vf, then obtain the expression for the change in the quantity1/2 m v^2. See if the result is equal to Fnet * `ds.
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Your solution:
1/2mv^2 : ½ mvf^2 - ½ mv0^2
= ½ m(v0^2 +2 Fnet/m *ds) - ½ mv0^2
= ½ mv0^2 + 1/2 m* 2 Fnet/m * ds - ½ mv0^2
= ½ m * 2 Fnet/m *ds
= Fnet/ds
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Question: `q007. We call the quantity 1/2 m v^2 the Kinetic Energy, often abbreviated KE, of the object.
We call the quantity Fnet * `ds the work done by the net force, often abbreviated here as `dWnet.
Show that for a net force of 12 Newtons and a mass of 48 kg, the work done by the net force in accelerating an object from rest through a displacement of 100 meters is equal to the change in the KE of the mass.
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Your solution:
Fnet = 12 Newtons, mass = 48 kg, ds =100 m
a = 12 Newtons/48 kg
a = .35 m/s^2
vf = +/- sqrt(2(.25 m/s^2)(100 m))
vf = +/- sqrt(50 m^2/s^2)
vf = +/- 7.07 m/s = +/- 7.1 m/s
KEf = ½(48 kg)(7.1 m/s)^2
KEf = 1209.84 J
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Question: `q008. How much work is done by the net force when an object of mass 200 kg is accelerated from 5 m/s to 10 m/s? Find your answer without using the equations of motion.
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Your solution:
KE0 = ½(200 kg)(5 m/s)^2
KE0 = 2500 J
KEf = ½(200 kg)(10 m/s)^2
KEf = 10,000 J
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Question: `q009. Answer the following without using the equations of uniformly accelerated motion:
If the 200 kg object in the preceding problem is uniformly accelerated from 5 m/s to 10 m/s while traveling 50 meters, then what net force was acting on the object?
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Your solution:
ds = 50 m
dWnet /ds = Fnet * ds/ds
dWnet/ds = Fnet
7500 Newtons/50 m = Fnet
150 Newtons = Fnet
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Question: `q010. Solve the following without using any of the equations of motion.
A net force of 5,000 Newtons acts on an automobile of mass 2,000 kg, initially at rest, through a displacement of 80 meters, with the force always acting parallel to the direction of motion. What velocity does the automobile obtain?
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Your solution:
dWnet = 5000 Newtons * 80 m
dWnet = 400,000 J
vf = +/- sqrt((2*400,000 J)/(2000 kg))
vf = +/- sqrt(400 J/kg)
vf = +/- 20 m/s
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Question: `q011. If the same net force was exerted on the same mass through the same displacement as in the previous example, but with initial velocity 15 m/s, what would then be the final velocity of the object?
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Your solution:
Fnet = 400,000 J
KE0 = ½ mv0^2
KE0 = ½(2000 kg)(15 m/s)^2
KE0 = 225,000 J
KEf = KE0 + dKE
KEf = 225,000 J + 400,000 J
KEf = 625,000 J
vf = +/- sqrt(2*625,000 J /2000 kg)
vf = +/- sqrt(625 m^2/s^2)
vf = +/- 25 m/s
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Question: `q012. Solve without using the equations of motion:
A force of 300 Newtons is applied in the direction of motion to a 20 kg block as it slides 30 meters across a floor, starting from rest, moving against a frictional force of 100 Newtons.
How much work is done by the net force, how much work is done by friction and how much work is done by the applied force?
What will be the final velocity of the block?
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Your solution:
Fnet = 300 N - 100 N
Fnet = 200 N
dWnet = 200 N * 30 m
dWnet = 6000 J
dWapplied = 300 N * 30 m
dWapplied = 9000 J
dWfrict = -100 N * 30 m
dWfrict = -3000 J
9000 J - 3000 J
= 6000 J
vf = +/- sqrt(2(6000 J)/(20 kg))
vf = +/- sqrt(12,000 J/20 kg)
vf = +/- sqrt(600 m^2/s^2)
vf = +/- 24.5 m/s
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If you understand the assignment and were able to solve the previously given problems from your worksheets, you should be able to complete most of the following problems quickly and easily. If you experience difficulty with some of these problems, you will be given notes and we will work to resolve difficulties.
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Question: `q013. What is the velocity of a 5 kg mass whose kinetic energy is 50 Joules?
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Your solution:
vf = +/- sqrt(50 J/5 kg)
vf = +/- sqrt(10 m^2/s^2)
vf = +/- 3.16 m/s
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Question: `q014. A 1500 kg automobile is moving at 10 m/s.
If its kinetic energy increases by 100 000 Joules, how fast will it be moving?
Suppose the automobile accelerated uniformly as it gained the 100 000 Joules of kinetic energy, which it gained in 10 seconds. What can be determined from this information? Indicate all the answers you can to this question, and how these quantities can be determined.
(University Physics students): If the energy is added at a constant rate of 10 000 Joules / second, will the distance traveled by the automobile be the same as, greater than or less than the distance calculated previously. Consider the fact that a constant force will not add energy at a constant rate to an object whose velocity is changing.
Challenging question: If the distance is different how far will the automobile travel during that time?
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Your solution:
a. 100,000 J/15,000 kg*m/s
= 6.66 m/s
@&
Dividing the increase in KE by the momentum does not give the velocity.
*@
b.
c.15,000 kg*m/s/100,000 J
= .66 m
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... nor does dividing the momentum by the change in KE give us the displacement.
*@
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Self-critique (if necessary):
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&#Good responses. See my notes and let me know if you have questions. &#