Conservation of Energy Example 2

Guide 8-6b. Conservation of Energy Example 2

This is an example of a connected object problem that can be solved using conservation of energy. The problem situation is shown in this applet.

Given: m₁ = 2.0 kg m₂ = 1.6 kg μ_k = 0.20 v_i = 0 θ = 30° We can determine the distance d that block 2 falls by reading positions on the applet. From this, d = 4.70 - 0.47 m = 4.23 m. Goal: Find the velocity of block 2 just before it touches the black platform. System: We'll take the two blocks and the Earth as the system. We don't include the pulley, because we're assuming it has no mass. External forces: The normal force of the plane does no work. The kinetic friction force does work on block 1; hence, we'll have a non-zero W_ext term. The string and tension forces (see diagram to right) do work of equal magnitude but opposite sign so that the net work due to tension is 0. initial state = both blocks motionless before block 2 is released final state = block 2 just reaches the black platform	This diagram shows that the total work done by the tension forces is zero.
Now let's look at the energy changes. The kinetic energy of the system of the two blocks obviously increases. The gravitational potential energy of block 2 decreases but that of block 1 increases, since the former block falls and the latter rises. In order to determine the net change in gravitational potential energy, examine the figure to the right. We pick the lowest point of block 2 as y = 0 and the positive direction as up. Note that the vertical distance d that block 2 falls is the same as the distance that block 1 moves along the plane. We can say this, because the string doesn't stretch. The changes in elevation of the two blocks are the following. y_2f - y_2i = 0 - d = -d y_1f - y_1i = dsinθ Therefore, the net change in gravitational potential energy is .
Now let's look at the work done by the friction force. We need to analyze the forces on block 1 in order to determine the friction force. See the figure to the right. We use a net force equation to find the normal force.
We use the previous result to find the friction force, and we use the definition of work to find the work done by the friction force. The force-displacement diagram is shown. Putting it all together.
Now we're ready to employ conservation of energy. We substitute the previous results for the work done by friction and the net change in gravitational potential energy. We also substitute for the kinetic energy change. Note that for the latter, we use the sum of the masses, since both blocks move at the same velocity at all times. Substituting v_i = 0 and rearranging, Finally, we solve for the final velocity. We select the negative root due to the way we defined the direction of positive vertical displacement. Checks are shown in the right-hand column.	Checks: In order to see if the result for final velocity makes sense, let's look at the problem a different way. We know that for a uniformly-accelerating object (or system in this case) that the following is true. Solving for the final velocity for the situation of interest, we obtain . By substituting the acceleration for various situations, we can see if we obtain expected results. Suppose the plane is frictionless. Then μ_k = 0. Recall the solution of this problem here in which the acceleration was found to be the following. Substituting this result into gives the same result as substituting μ_k = 0 into the result for v_f in the left-hand column.