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 |
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.
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,
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.