Energy Bar Chart Simulator

A block starts from rest at the top of a ramp whose top is 5 m above the ground, and slides down. The left panel shows the ramp and where the block is; the right panel is a bar chart with four bars: KE (kinetic), PE (gravitational potential), Therm (thermal energy produced by friction) and Total. The simulator uses $g = 9.8$ m/s² and measures PE from the ground.

The readout lists KE, the block's speed v, PE, Thermal and Total, each to 0.1, with a note underneath. With μ = 0 the note says KE + PE is conserved; try changing the angle and notice that the speed at a given height does not change. With friction on, the note says KE + PE + thermal equals the total.

If friction is too strong for the angle ($\mu \ge \tan\theta$), the note warns that a released block would not slide at all. In that case it would stay at the top, so the bars at lower positions no longer describe a real motion; lower μ or steepen the ramp before reading them. With μ = 0.60, for example, the block only slides on ramps steeper than about 31°.

On the AP Physics 1 exam, kinetic energy is written $K$ and gravitational potential energy $U_g$: $$K = \tfrac{1}{2}mv^2 \qquad \Delta U_g = mg\Delta y$$

For the system of the block, the Earth and the ramp, no external work is done, so the total energy is constant: $$K_i + U_{g,i} = K_f + U_{g,f} + \Delta E_{th}$$ With no friction, $\Delta E_{th} = 0$ and the energy lost from $U_g$ all appears as $K$.

Friction converts mechanical energy into thermal energy. On a ramp the friction force is $f = \mu mg\cos\theta$, and the block slides a distance $d = \Delta h/\sin\theta$ along the surface while dropping $\Delta h$, so $$\Delta E_{th} = f d = \frac{\mu}{\tan\theta}\, mg\Delta h$$ A shallower ramp means a longer path, so more energy goes to thermal for the same drop.

If you instead choose the block and the Earth as the system, friction is an external force doing negative work, $W = -fd$, and the system's mechanical energy decreases by that amount. Both descriptions give the same speed; what changes is which bars belong to the system.

A 2 kg block starts from rest at the top of a 30° ramp (5 m high) with $\mu = 0.20$. Find its energies and speed when it has dropped to a height of 2.0 m.

Total. $E = mgh_{top} = (2)(9.8)(5) = 98.0$ J, all of it $U_g$ at the start.

Potential energy at 2.0 m. $U_g = (2)(9.8)(2.0) = 39.2$ J. The block has dropped 3.0 m, releasing $(2)(9.8)(3.0) = 58.8$ J.

Thermal energy. The path length is $3.0/\sin 30° = 6.0$ m and the friction force is $(0.20)(2)(9.8)\cos 30° = 3.39$ N, so $\Delta E_{th} = (3.39)(6.0) = 20.4$ J.

Kinetic energy and speed. $K = 58.8 - 20.4 = 38.4$ J, so $v = \sqrt{2(38.4)/2} = 6.2$ m/s. Check: 38.4 + 39.2 + 20.4 = 98.0 J.

In the simulator, set θ = 30°, m = 2 kg, μ = 0.20 and drag the position to 2.0 m. The readout should match: KE 38.4 J, v 6.2 m/s, PE 39.2 J, Thermal 20.4 J, Total 98.0 J. Set μ back to 0 and the speed at the same height rises to 7.7 m/s, because all 58.8 J becomes kinetic energy.

Energy bar charts appear throughout Unit 3 (Topic 3.4, Conservation of Energy) and return in Units 6 and 7 for rotating objects and oscillators. Free-response questions often ask you to complete a bar chart for a second position and then use it to write an energy equation, so make the bar heights consistent with the equation you write.