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Is mechanical energy conserved when moving down an incline? If so, under what conditions is all of the potential energy converted to kinetic energy, leading to the greatest speed at the bottom?
According to the law of conservation of energy, in an isolated system (frictionless), all of the gravitational potential energy of an object at the top of an incline should become kinetic energy when that object reaches the bottom of the incline. In a non-isolated system, when friction is present, energy will be lost; therefore, the total mechanical energy will decrease over time as the amount of potential energy available to be converted to kinetic energy decreases. The applet used in the following simulation helps you explore the motion and forces acting on a mass moving down a smooth, frictionless plane.
Open the Inclined Plane, Frictionless simulation. You can learn more about the simulation and how to use it by reading the Show Me found at the top of the simulation screen.
On the simulation, select the “View Graph” tool (). Then complete the following steps:
Check to ensure that the coefficient of friction is set to zero.
Press “Play,” and watch the changes in potential and kinetic energy. On the graph, select the “Fit graph” tool to make the graph fully visible. Drag the margins to the right and down to enlarge the window. Click on the down arrow just under the “Zoom In” button to expand the graph further downward.
SC 1. Sketch the graph for potential energy, kinetic energy, and mechanical energy versus time.
SC 1.
Remember to submit the answers to LAB 1 and LAB 2 to your teacher as part of your Module 6: Lesson 2 Assignment.
LAB 1.
LAB 2. Assume that the block has a potential energy of 0 J when it reaches the bottom of the incline. What can be said about the sum (mechanical energy) of the potential and kinetic energy for the block at any point on the way down the incline when no friction is present?
Read “Mechanical Energy” on pages 306 to 309 of your textbook.
SC 2. Complete question 2 of “Practice Problems” on page 309 of your textbook.
SC 2.
m = 4.50 kg
Em = 6.27 × 104 J
h = 275 m
the speed of the cannon ball (v)
The cannon ball will possess potential energy because of its height and kinetic energy as a result of its speed. Find speed from the kinetic energy using the total mechanical energy formula.
Em = Ek + Ep
Ek = Em – Ep
The speed of the cannon ball is 150 m/s.
TR 1. Solve using energy considerations. Recall from the previous lesson that Ep = ½ kx2. Since F = kx, Ep = ½ Fx.
Read about mechanical energy in isolated systems on pages 311 to 314 of your textbook.
SC 3. Complete question 1 of “Practice Problems” on page 313 of the textbook.
SC 3.
Eki = 250 J
ΔEp = 650 J (loss)
the final kinetic energy (Ekf)
The ramp is frictionless, so this is an isolated system. According to the law of conservation of energy, the loss of potential energy will equal the gain in kinetic energy. Therefore, the final kinetic energy will equal the initial kinetic energy less the change in potential energy.
ΔEk = – Δ Ep
Ekf – Eki = – Δ Ep
Ekf = Eki – Δ Ep
Ekf = (250 J) – (–650 J)
Ekf = 900 J
The final kinetic energy of the crate is 900 J.
Remember to submit the answer to TR 2 to your teacher as part of your Module 6: Lesson 2 Assignment.
TR 2. Solve these questions using energy considerations.
a. A 0.80-kg coconut is growing 10 m above the ground in its palm tree. The tree is just at the edge of a cliff that is 15 m tall. What would the maximum speed of the coconut be if it fell to the ground beneath the tree?
b. What would the maximum speed be if it fell from the tree to the bottom of the cliff?
Read about mechanical energy in pendulums on pages 314 to 316 of the textbook.
SC 4. Complete question 1 of “Practice Problems” on page 315 of the textbook.
SC 4.
hf = 15.0 cm
hi = 25.0 cm
m = 250 g
(a) the mechanical energy (Em)
(b) the kinetic energy (Ekf)
(c) the speed (v)
(a) The pendulum is frictionless. So this is an isolated system, and mechanical energy will be conserved. The mechanical energy at any point is equal to the initial potential energy because there was no kinetic energy at the start.
Emi = Epi + Eki
Emi = mghi + ½mvi2
Emi =(0.250 kg)(9.81 m/s2)(0.250 m) + ½(0.250 kg)(0 m/s)2
Emi = 0.613 J
(b) According to the law of conservation of energy, the loss of potential energy will equal the gain in kinetic energy. Therefore, the kinetic energy will equal the initial mechanical energy less the change in potential energy.
Ekf = Emf – ΔEpi
Ekf = Emf –mgΔh
Ekf = (0.613 J) –(0.250 kg)(9.81 m/s2)(0.250 m – 0.15 m)
Ekf = 0.245 J
(c) Use the kinetic energy formula to find the speed.
Ekf = ½mvf2
(a) The mechanical energy is 0.613 J.
(b) The kinetic energy is 0.245 J.
(c) The speed is 1.40 m/s.
Remember to submit the answer to TR 3 to your teacher as part of your Module 6: Lesson 2 Assignment.
TR 3. A pendulum is dropped from the position as shown in the diagram (0.50 m) above the equilibrium position. What is the speed of the pendulum bob as it passes through the equilibrium position?
Up to now in this lesson, we have dealt with a frictionless environment. What happens when we consider mechanical energy systems that include friction? Complete the rest of the lab to see.
Remember to submit the answer to LAB 3 to your teacher as part of your Module 6: Lesson 2 Assignment.
LAB 3. If necessary, re-open the Inclined Plane, Frictionless simulation.
Set the coefficient of friction to 0.2, and repeat the procedure. You may have to click “Zoom Out” several times to see the whole graph. Sketch the three graphs on the grid below. Label the lines and axes.
Repeat several trials with greater coefficients of friction each time, up to a maximum of 0.55. Observe how the mechanical energy changes in each case.
SC 5. Describe the relationship between the coefficient of friction and the total mechanical energy as the object slides down the incline.
SC 5. As the coefficient of friction increases, the total mechanical energy as the object slides down the incline decreases.
Remember to submit the answers to LAB 4 and LAB 5 to your teacher as part of your Module 6: Lesson 2 Assignment.
LAB 4.
LAB 5. Assume that the block has a potential energy of 0 J when it reaches the bottom of the incline.
The total mechanical energy (sum of the potential and kinetic energy) is conserved in an isolated environment. Any change in either potential or kinetic energy is associated with an equal but opposite change in the other energy. As an object slides down an incline, potential energy is converted into kinetic energy until all of the energy has been transferred from one form to the other.
Mechanical energy is not conserved when friction is present, making the system non-isolated. In this case, some of the potential energy is converted to heat due to the friction, rather than being converted to kinetic energy. Therefore, the mechanical energy decreases over time. By reducing the friction, the loss in mechanical energy can be minimized, resulting in a greater conversion of potential energy into kinetic energy and a greater speed at the bottom of an incline.
conservative force: a force such as gravity, acting in an isolated system, where the total work done is independent of the path an object is moved through
non-conservative force: a force acting on a non-isolated system from outside the system or from friction; a force where the total work done depends on the path an object is moved through
It is important to note that the forces related to potential energy (gravitational and elastic forces) and kinetic energy act within the isolated system and do not affect the mechanical energy of the system. These forces are called conservative forces. Other forces, such as friction and forces from outside a non-isolated system, affect the mechanical energy of the system and are called non-conservative forces. The following Conservative and Non-conservative Forces simulation gives a visual and graphical representation of these forces.