Conservation of Energy

The skier is in a conservative system, as the only force acting upon him is gravity. Instead of calculating the work done over the curved hills, we can construct an alternate path, because of the principle of path independence:

We construct a path of two segments: one is horizontal, going between the two hills, and one is vertical, accounting for the vertical drop between the two hills. What is the work done over each of these two segments? Since the gravitational force is perpendicular to the displacement in the horizontal segment, no work is done. For the second segment, the gravitational force is constant and parallel to the displacement. Thus the work done is: W = Fx = mgh = 10mg . By the Work-Energy Theorem, this net work causes an increase in velocity. If the skier started with no initial velocity, then we can relate the final velocity to the work done:

We can cancel the mass and solve for v _f :

Thus the final velocity of the skier is 14 m/s.

Remember that ΔU = - W . We had calculated that the gravitational force exerted a work of 10mg during the entire trip. Thus the change in potential energy is simply the negative of this quantity: ΔU = - 10mg = - 500g = - 4900 Joules. The potential energy lost is converted into kinetic energy, accounting for the final velocity of the skier.

Here we have a system of two conservative forces, mass and gravity. Even if there are more than one conservative force acting in a system, it is still a conservative system. Thus potential energy is defined, and we can calculate the total energy of the system. Since this quantity is constant, we may choose any position for the mass that we like. In order to avoid calculating kinetic energy, we choose a point at which the mass has no velocity: at its maximum displacement, the position shown in the figure above. Also, since energy is relative, we may choose our origin to be the equilibrium point of the spring, as shown in the figure. Thus both the gravitational force and the spring force contribute to the potential energy: U _G = mgh = - 5mg = - 245 Joules. Also, U _s = kx ² = (10)(5)² = 125 Joules. Thus the total potential energy, and hence the total energy is the sum of these two quantities: E = U _G + U _s = - 120 Joules. Remember that answers may vary on this problem. If we had chosen a different origin for our calculations, we would have gotten a different answer. Once we have chosen an origin, however, the answer for total energy must remain constant.

We know that if the particle completes a closed path, the net work on the particle is zero. We already established through the Work-Energy Theorem that the total kinetic energy does not change. However, we also know that ΔU = - W . Since no work is done, the potential energy of the system does not change.

We can also answer this question in a more conceptual manner. We have defined potential energy as the energy of configuration of a system. If our particle returns to its initial position, the configuration of the system is the same, and must have the same potential energy.

In this case there are two forces acting on the ball: gravity and tension from the spring. The tension, however, always acts perpendicular to the motion of the ball, thus contributing no work to the system. Thus the system is a conservative one, with the only work being done by gravity. When the pendulum is elevated, it has a potential energy, according to its height above its lowest position. We can calculate this height:

The height h can be calculated by subtracting x from the total length of the string: h = 1 - x . We use a trigonometric relation to find x: sin30 ^o = . Thus x = .5m and h = 1 - .5 = .5m . Now that we have the initial height of the pendulum, we can calculate its gravitational potential energy: U _G = mgh = .5mg . All of this potential energy is converted into kinetic energy at the final position of the pendulum, with a height of 0. Thus: .5mg = mv ² . The masses cancel, and we can solve for v: v = = 3.1m/s . Thus, when the pendulum reaches an angle of 90 with the horizontal, it has a velocity of 3.1 m/s.