query 12

012. `query 12

Physics I

08-02-2008

......!!!!!!!!...................................

22:13:36

Query set 3 #'s 13-14 If an object of mass m1 rests on a frictionless tabletop and a mass m2 hangs over a good pulley by a string attached to the first object, then what forces act on the two-mass system and what is the net force on the system? What would be the acceleration of the system? How much would gravitational PE change if the hanging mass descended a distance `dy?

......!!!!!!!!...................................

RESPONSE -->

The net force for m1 is the mass of m1 * 9.8m/s^2 * a slope of 0. Therefore the net force of m1 is 0.

The net force of m2 is the mass of m2 *9.8m/s^2.

Add those together, or rather since m1 = 0 , the net force of the system is just m2*9.8m/s^2.

To find acceleration divide the net force of the system, as stated above, by the total mass of the system (m1 + m2).

I think the gravitational PE would decrease by the product of Fnet (m2*9.8m/s^2)* distance dy. But I'm not sure.

confidence assessment: 1

.................................................

......!!!!!!!!...................................

22:14:35

** The net force on the system is the force of gravity on the suspended weight: Fnet = m2*9.8m/s/s

Gravity also acts on m1 which is balanced by force of table on m1, so this force makes no contribution to Fnet.

Acceleration=net force/total mass = 9.8 m/s^2 * m2 / (m1+m2).

If the mass m2 descends distance `dy then gravitational PE decreases by - m2 g * `dy.

COMMON MISCONCEPTIONS AND INSTRUCTOR COMMENTS:

The forces acting on the system are the forces which keep the mass on the table, the tension in the string joining the two masses, and the weight of the suspended mass. The net force should be the suspended mass * accel due to gravity + Tension.

INSTRUCTOR COMMENT:

String tension shouldn't be counted among the forces contributing to the net force on the system.

The string tension is internal to the two-mass system. It doesn't act on the system but within the system.

Net force is therefore suspended mass * accel due to gravity only

'The forces which keep the mass on the table' is too vague and probably not appropriate in any case. Gravity pulls down, slightly bending the table, which response with an elastic force that exactly balances the gravitational force. **

......!!!!!!!!...................................

RESPONSE -->

self critique assessment: 3

.................................................

......!!!!!!!!...................................

22:25:12

How would friction change your answers to the preceding question?

......!!!!!!!!...................................

RESPONSE -->

When finding the net force the force of friction has to be added in. We already have the net force of m1 = 0, and the net force of m2 = m2*9.8m/s^2. To calculate ffrict first find the normal force of m1 (m1*9.8m/s^2) and multiply it by the coefficient of friction ( a given percentage of weight of m1). Because the friction acts in opposition to the direction of the movement subtract ffrict from the net force previously calculated to find the new net force. Net force will be lower. Then proceed to calculate acceleration, etc.

confidence assessment: 1

.................................................

......!!!!!!!!...................................

22:25:32

**Friction would act to oppose the motion of the mass m1 as it slides across the table, so the net force would be m2 * g - frictional resistance. **

......!!!!!!!!...................................

RESPONSE -->

self critique assessment: 3

.................................................

......!!!!!!!!...................................

22:28:27

Explain how you use a graph of force vs. stretch for a rubber band to determine the elastic potential energy stored at a given stretch.

......!!!!!!!!...................................

RESPONSE -->

The trapezoidal area under the curve = the total work done to stretch the rubber band. Since the work done is conservable it also equals the PE. (Except my notes say something about neglecting thermal losses, so I am guessing it isn't quite exact, but close enough for me at this time.)

confidence assessment: 2

.................................................

......!!!!!!!!...................................

22:29:22

** If we ignore thermal effects, which you should note are in fact significant with rubber bands and cannot in practice be ignored if we want very accurate results, PE is the work required to stretch the rubber band. This work is the sum of all F * `ds contributions from small increments `ds from the initial to the final position. These contributions are represented by the areas of narrow trapezoids on a graph of F vs. stretch. As the trapezoids get thinner and thinner, the total area of these trapezoids approaches, the area under the curve between the two stretches.

So the PE stored is the area under the graph of force vs. stretch. **

......!!!!!!!!...................................

RESPONSE -->

self critique assessment: 3

.................................................

......!!!!!!!!...................................

22:37:20

STUDENT QUESTIONS: Does the slope of the F vs stretch graph represent something? Does the area under the curve represent the work done? If so, is it work done BY or work done ON the rbber bands?

......!!!!!!!!...................................

RESPONSE -->

The area under the curve represents the work done ON the rubber band, but I cant think of what the slope might represent.

confidence assessment: 1

.................................................

......!!!!!!!!...................................

22:38:55

** Slope isn't directly related to any physical quantity. The area is indeed with work done (work is integral of force with respect to displacement).

If the rubber band pulls against an object as is returns to equilibrium then the force it exerts is in the direction of motion and it therefore does positive work on the object as the object does negative work on it.

If an object stretches the rubber band then it exerts a force on the rubber band in the direction of the rubber band's displacement, and the object does positive work on the rubber band, while the rubber band does negative work on it. **

......!!!!!!!!...................................

RESPONSE -->

So the area under the curve can be either the work done BY or the work done ON, depending on what the graph is actually representing?

self critique assessment: 1

The present graph represents the work done on the rubber band, and the negative of the work done by the rubber band on the object.

The meaning of the graph depends on interpretation of the quantities and, as you say, on what the graph represents.

.................................................

......!!!!!!!!...................................

......!!!!!!!!...................................

Tension in a stretched object results in forces throughout the object.

The tension itself has no direction, but it exerts equal and opposite forces at every point of the rubber band except at the ends.

At each end the tension force toward the other end is unopposed by a tension in the opposite direction (e.g., at the right end there is a tension force toward the left but, unlike at a point between the two ends, there is no equal and opposite force to the right). So at each end the tension exerts a force toward the other end.

142.9 N * cm is pretty close to 1.66 J.

A Joule is a N * m, and it takes 100 cm to make a meter. So 142.9 N * cm = 142.9N * (.01 m) = 1.429 N m = 1.429 Joules.

If you include the point (0, 0), which is reasonable for the very reasons you give, the graph is concave down. The region beneath the curve is a bit greater than that of the the single trapezoidal region you calculated, which accounts for the rest of the difference. If you had used two trapezoids instead of one, your answer would have been close to the given answer.

.................................................

&#This looks good. See my notes. Let me know if you have any questions. &#