68 Wheeling a Patient

Thermal Energy

In the previous chapter we analyzed the stopping distance for objects brought to rest by friction and claimed that friction did work to transfer  kinetic energy into thermal energy. Thermal energy is not actually a new type of energy, but rather just kinetic energy on a microscopic scale. Thermal energy is the energy stored in the motion of atoms and molecules that make up a material. Transferring thermal energy to a system really just means that you caused it’s atoms and molecules to move faster. You can visualize how friction can transfer kinetic energy to thermal energy in the simulation below. Work done to permanently deform a material (into the failure region of the strain curve) also transfers energy to thermal energy.

Reinforcement Exercises

Multiple Works

The systems in the previous chapters only experienced one external force (either friction or gravity). When a system experiences multiple external forces we can use W = Fdcos\theta to calculate the work done by each force and then add up the works (keeping negative works as negative) to get the net work. Alternatively, we can add up the forces as vectors (keeping track of directions) to find the size and direction of the net forcey] and then use the net force in the work equation: W_{net} = F{net}dcos\theta.  Either way will give you the same answer, which will be the net work.

Everyday Examples: Moving a Gurney

To move a 80 kg patient to another room, Jolene wants to get the patient and the 50 kg gurney they lie on up to a speed of 0.8 m/s and maintain that speed. The old wheels on the gurney produce 20 N of friction that resists rolling. Jolene can comfortably apply a 100 N horizontal force to the gurney.

  1. For what distance should she apply this 100 N force to get the gurney up to 0.5 m/s?
  2. How far from her destination should she stop pushing the gurney to allow friction to stop it at the right place?

Our system will be the patient and gurney, but NOT the wheels. That way the frictional force and the force from Jolene are both external to the system. The net work done by these forces will contribute to changing the kinetic energy in the center of mass motion of the system.

Once up to speed Jolene will need to push with that same size of force that friction provides until she gets to her designation. That will ensure that net force and net work will be zero and the speed of the gurney won’t change during that part of the trip. However, we need to find the distance to apply the 300 N and get up to speed.

The work-energy principle tells us to set the change in kinetic energy equal to the net work.

(1)   \begin{equation*} F_{net}dcos\theta = \frac{1}{2}mv_f^2-\frac{1}{2}mv_i^2 \end{equation*}

We will keep in mind that the gurney started at rest, so the initial kinetic energy was zero.

(2)   \begin{equation*} F_{net}dcos\theta = \frac{1}{2}mv_f^2-0 \end{equation*}

The gurney will move horizontally and the net force is also horizontal so the angle between net force and motion is zero and cos 0 = 1. (We know the angle was not 180° because the kinetic energy of the gurney is increasing so the net work is positive and it must be moving in the same direction as the net force).

(3)   \begin{equation*} F_{net}d = \frac{1}{2}mv_f^2 \end{equation*}

We can enter values to calculate the final kinetic energy, keeping in mind the total mass of the gurney and patient is 130 kg.

(4)   \begin{equation*} F_{net}d = \frac{1}{2}130\, \bold{N} (0.8 \, \bold{m/s}) ^2 \end{equation*}

After entering the values for mass and speed we get:

(5)   \begin{equation*} F_{net}d = 41.6\, \bold{J} \end{equation*}

Now we just need to calculate the net force that results from the push and friction forces:

(6)   \begin{equation*} F_{net} = 100\, \bold{N} - 20\, \bold{N} = 80\, \bold{N} \end{equation*}

And enter that into our work-energy equation:

(7)   \begin{equation*} 80\, \bold{N}(d) = 16.25\, \bold{J} \end{equation*}

Then divide by that 80 N net force to isolate d:

(8)   \begin{equation*} d = \frac{41.6\, \bold{J}}{80\, \bold{N}} = 0.5 \bold{m} \end{equation*}

We find that Jolene should apply 100 N of force for about half a meter to get up to speed. Then she should drop her applied force down to 20 in order to maintain speed.

The 20 N rolling resistance caused by friction does the work of stopping gurney. To find necessary distance we use the work-energy theorem again, knowing that gurney now has 41.6 J of kinetic energy and zero when comes to a stop:

(9)   \begin{equation*} F_{net}dcos\theta = 0-41.6\, \bold{J} \end{equation*}

We  know that the net force is just the 20 N rolling resistance acting 180° to the direction of motion (backwards). We can calculate that cos\180^{\circ} =-1 so we have:

(10)   \begin{equation*} (20\, \bold{N})d(-1) = -41.6\, \bold{J} \end{equation*}

Solving for d:

(11)   \begin{equation*} d = \frac{41.6\, \bold{J}}{(20\, \bold{N})} = 2.08\, \bold{m} \end{equation*}

Jolene will need to let the gurney stop rolling about 2 m from her destination if she wants to let friction do the work and not pull back on it at all.

Remembering that energy must be transferred when work is done, we might ask where the kinetic energy went as the gurney in the previous example came to rest. The answer is thermal energy in the wheels, which were not part of our system. If we wanted to include the wheels in our system then we need to be ready to analyze systems with changing thermal energy, which we will do soon.

Reinforcement Exercises



Icon for the Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License

Body Physics 2.0 Copyright © by Lawrence Davis is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted.

Share This Book