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تاريخ التسجيل : 21/09/2008
 |  موضوع: potential energy  2008-10-12, 6:02 am | |
| Applications of Work-Energy : Potential EnergyIn order to continue the idea of conservation, we note that the work done by gravity has a property which is distinct from work done by friction. Namely, if we lift, with constant velocity, an object in a gravitational field, and then lower it to its original position, the total work done is zero since work depends only on the change in vertical height. We call such forces conservative. Friction is not conservative since if we drag a block on any surface with friction, the work done depends only on how far we dragged it, not where the path of travel began and ended. Conservative forces have the nice property of being able to be defined in terms of a potential energy. The usual definition of potential energy is through the work-energy theorem as for kinetic energy, i.e. W = Ui - Uf. To see that this is a useful way of thinking about things, consider again the case of an object thrown straight up with an initial velocity v0. The object starts with kinetic energy K0 = ½ mv02 and ends at the maximum height with Kf = 0. Given the work done by gravity (negative in this case), our definition of potential energy would say that <td align=right>-DU</FONT> | = W ==> | <td align=right>-DU</FONT> | = DK</FONT> |
So we can consider the "loss" of kinetic energy as going into a "gain" of potential energy. As the object falls, the potential energy is transformed back into kinetic energy. We can define a change in potential energy also through the action of a force over a distance. The potential energy is useful in that it keeps us from calculating the work done by a force when the path is complicated, or tedious to follow. Since the work done depends only on the initial and final positions, the change in potential (and kinetic) energy depends only on change in position as well. Through the work energy theorem, we have K f - K i = W = U i - U f ==> K i + U i = K f + U f = constant = E where E is called the total mechanical energy. Here's an example of its use with another concept, uniform circular motion.
- Example 1:
A 5 kg mass on a horizontal table is connected by an ideal string over an ideal pulley to a 3 kg mass that hangs freely as shown in the figure.
The system is held at rest and released at time t = 0. You make measurements that show that when the 3.00 kg mass has fallen thru 0.80 m, the speed of the 5.00 kg mass is 1.50 m/s. Your task is to determine whether there is friction between the table and the 5 kg mass and if so what the coefficient of kinetic friction must be.
Solution:
This is an example of a problem that you can solve using Newton's Laws, but let's instead consider the work-energy theorem to show the advantage of this approach in terms of cutting short the calculation.
We can save some time here (and in many problems involving work-energy) by choosing the origins of the coordinate system carefully. In this case, let's set the origin of the coordinate system as having y = 0 at the original position of the 3.00 kg mass and x = 0 at the starting position of the 5.00 kg mass. Then the initial kinetic and potential energy of the system consisting of both masses is zero and the initial mechanical energy of the system is thus zero. So,
| Kf + Uf + W | = 0 | | ½ (5.00 kg + 3.00 kg)(1.50 m/s)2 + (3.00 kg)(9.8 m/s2)(-0.80 m) + W | = 0 ==> | | W | = 14.5 J |
This shows that friction definitely was acting since 14.5 J of energy was ``stolen''. The masses would have been moving faster otherwise. To find the coefficient of kinetic friction we look at the work done by friction
| Wf | = 14.5 J | | fk(0.80 m) | = 14.5 J ==> | | mk (5.00 kg)(9.8 m/s2) | = 18.1 Nt ==> | | mk | = 0.37 |
- Example 2:
The figure above shows a pendulum with a small bob of mass m at the end of a massless, stretchless string of length L. A peg is located 2/3 L directly underneath the point of suspension for the string. What initial angle, q0, must the pendulum be released from so that it just completes a loop-the-loop?
Solution: The bob must swing through an arc of length L until the string reaches the peg. At that point, the string wraps around the peg so that the bob swings in an arc of radius 1/3 L instead of L. To just complete a loop-the-loop, the bob needs to maintain circular motion by having the net force acting on it always be equal to mv2/(1/3 L). Since we know that the potential energy goes up and the kinetic energy goes down as the bob rises vertically, the lowest kinetic energy (and hence the slowest speed) occurs when the bob is directly above the peg. Since the bob starts with zero kinetic energy, the kinetic energy when it is above the peg comes from the difference in potential energy given its initial height (when we released it) and its height when it is above the peg. To measure these heights, we need a common origin, so let's choose the bottom of the pendulum's swing. The initial height is then, L - L cos(q0) = L (1 - cos(q0)). The height when the bob is directly above the peg is twice the radius of the arc after the string hits the peg. This is 2(1/3 L) = 2L/3. Therefore, the kinetic energy is
| Kf - K0 | = U0 - Uf | | ½ mv2 - 0 | = mgL(1 - cos(q0)) - mg(2/3 L) = mgL(1/3 - cos(q0)) ==> | | v2 | = 2gL[1/3 - cos(q0)] ==> |
So, we have the velocity as a function of the initial angle the string makes with the vertical. To make a complete loop, we need to have the tension and weight satisfy the centripetal condition at all points along the loop. The tension can increase to any value needed for high speeds, but at low speeds, the tension is bounded by zero as its minimum value. Hence the minimum speed for the bob when it is at the top of the loop is T - mg = -mg = -mvmin2/(1/3 L) ==> vmin2 = gL/3. Substituting this minimum value into our expression for the velocity as a function of angle from the energy conservation equation leads to
| v2 | = vmin2 ==> | | 2gL[1/3 - cos(q0)] | = gL/3 ==> | | cos(q0) | = 1/3 ==> | | q0 | = cos-1(1/3) = 70.5° |
Finally, we define power as < P> = W/(t - t0) where the instantaneous power is P = dW/dt. Hence the integral of instantaneous power with respect to time is the work done. Power is measured in a number of different units, but the SI unit for power is the Watt, where 1 Watt = 1 J/s.
- Example 3:
A winch is used to lift a 200 kg mass to the roof of a 3 story building (a height of 10 meters). Assuming that the winch works at a steady rate of 200 Watts, how long will it take to lift the object to the roof. Ignore any frictional effects.
Solution: The winch supplies a mean power of 200 J/s. Assuming that we supply only a small amount of kinetic energy to keep the mass moving (i.e. we make the assumption that the time for lifting will be small enough that the velocity and hence the kinetic energy of the mass will be small compared to the change in potential energy), we can state that the total change in energy for the object is all a change in the potential energy. So,
Uf - U0 = mg(y - y0) = (200 kg)(9.8 m/s2)(10 m) = 19600 J
The mean power gives the rate at which this energy can be supplied so, (Uf - U0)/< P> = (19600 J)/(200 J/s) = 98 s. Let's check our original assumption about the size of the kinetic energy. Assuming a constant speed for the lift, v = (10 m)/(98 s) = 0.1 m/s ==> K = ½ (200 kg)(0.1 m/s) = 10 J. This is certainly negligible compared to the 19600 J the winch must supply to lift the object.
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