67 Work and Energy

Kinetic Energy

Movement is one of the the fundamental functions of the body. Like all the other function, motion involves energy. The amount of kinetic energy stored in the motion of the body, or any other system, depends on its and its speed:

(1)   \begin{equation*} KE = \frac{1}{2}mv^2 \end{equation*}

Reinforcement Activity

Work-Energy Theorem

The states that the total energy in an isolated system cannot change. In that case, changes to the kinetic energy of a system must be due to work was done on the system by an external force. The , or Work-Energy Theorem states that the net external work on a system is equal to the change in kinetic energy of the entire system.

(2)   \begin{equation*} W_{net} = \Delta KE \end{equation*}

The change in a quantity is always calculated as the initial value subtracted from the final value:

(3)   \begin{equation*} W_{net} = \frac{1}{2}mv_f^2-\frac{1}{2}mv_i^2 \end{equation*}

Examples: Falling Speed

A patient falls out of a bed that is 0.75 m above the floor. What speed does the person reach before hitting the floor (impact speed)?

When a person falls, gravity will be acting on them. If we want to calculate the fall speed, we can apply the work-energy theorem with the person as our system.

(4)   \begin{equation*} W_{net} = \Delta KE \end{equation*}

Neglecting air resistance, there  are no forces other than gravity on the person and gravity points in the same direction as the motion:

(5)   \begin{equation*} F_gdcos(0) = \Delta KE \end{equation*}

The weight can be calculated as F_g = mg

(6)   \begin{equation*} mgd = \Delta KE \end{equation*}

Then we write out the formula for kinetic energy and calculate the change as initial subtracted from final

(7)   \begin{equation*} mgd = \frac{1}{2}mv_f^2-\frac{1}{2}mv_i^2 \end{equation*}

The speed and initial kinetic energy at the start of the fall was zero.

(8)   \begin{equation*} mgd = \frac{1}{2}mv_f^2 \end{equation*}

We can divide every term in our equation by mass to cancel it out:

(9)   \begin{equation*} 2gd = \frac{1}{2}v_f^2 \end{equation*}

We just learned something interesting. The speed of an object in free-fall does not depend on its mass!

To continue we isolate the speed by multiplying both sides by 2 and taking a square root:

(10)   \begin{equation*} v_f = \sqrt(2gd) \end{equation*}

We are ready to calculate the final speed for the 0.75 fall.

(11)   \begin{equation*} v_f = \sqrt{2(9.8 ,\bold{m/s})(0.75,\bold{m})} = 3.8 ,\bold{m/s} \end{equation*}

The patient reaches an impact speed of roughly  4 m/s, or 9 mph.

Reinforcement Exercises

Everyday Examples

A person wearing socks is running at 4 m/s across a hardwood floor and then slides to a stop. The coefficient of between the socks and the floor is 0.4. How far will they slide before coming to a stop?

We can apply the work-energy theorem with the person as our system.

(12)   \begin{equation*} W_{net} = \Delta KE \end{equation*}

The work done on the system is due to kinetic friction. (The person is not moving vertically, and they don’t start moving vertically, so the vertical forces are in equilibrium and do no net work.)

(13)   \begin{equation*} F_{f,k} dcos\theta= \Delta KE \end{equation*}

The kinetic friction force can be calculated as F_{f,k} = \mu_k F_N. In this case the normal force is equal to the weight so we can write F_{f,k} = \mu_k mg. Friction acts opposite to the sliding motion so \theta = 180^{\circ}.

(14)   \begin{equation*} \mu_k mgd cos(180^{\circ})= \Delta KE \end{equation*}

Then we write out the formula for kinetic energy and calculate the change as initial subtracted from final

(15)   \begin{equation*} -\mu_k mgd  = \frac{1}{2}mv_f^2-\frac{1}{2}mv_i^2 \end{equation*}

The final speed and kinetic energy is zero when the person stops.

(16)   \begin{equation*} -\mu_k mgd = -\frac{1}{2}mv_i^2 \end{equation*}

We can divide both sides by mass to cancel it out. The negative on both sides also cancels out.

(17)   \begin{equation*} \mu_k g d  = \frac{1}{2}v_i^2 \end{equation*}

We just learned something interesting. The distance an object takes to stop does not depend on its mass!

To continue we isolate distance:

(18)   \begin{equation*} d =  \frac{1}{2}\frac{v_i^2}{\mu_k g} \end{equation*}

We are ready to enter our values and calculate the stopping distance:

(19)   \begin{equation*} d =  \frac{1}{2}\frac{(4\,\bold{m/s})^2}{0.15(9.8\,\bold{m/s/s}} =2\,\bold{m} \end{equation*}

The stopping distance is roughly 2 m.

Reinforcement Exercises

Everyday Example: Average Force During a Climbing Fall

A 65 kg rock climber is 4 m above their last anchor piece and falls. They will fall a total distance of 8 m before the rope begins to come tight and catch their fall.

  • If the rope, attached to their harness, stops them over a distance of 2 m, what is the average force on the climber from their harness?
  • If the climber were to land on their back on the sandy ground and stop in a distance of 2 cm, what is the average force on the climber from the ground?

We will apply the Work-Energy Theorem to the fall and the catch separately with the climber alone as our system.  This will allow us to calculate the average force applied to the climber during the catch.

(20)   \begin{equation*} W_{on} = \Delta KE \end{equation*}

Neglecting air resistance, there are no forces other than gravity on the climber and it points in the same direction as the motion:

(21)   \begin{equation*} F_gdcos(0) = \Delta KE \end{equation*}

The weight of the climber can be calculated as F_g = mg

(22)   \begin{equation*} mgd = \Delta KE \end{equation*}

To analyze the “catch” part of the fall we will again apply the work energy theorem, only now we know the climber must lose the same mgd of kinetic energy they just gained so now \Delta KE = -mgd. There are now two forces on the climber, weight (mg)and the unknown force from the harness F_h. Both forces act for the stopping distance (d_s), but the weight acts in the same direction as the motion during the catch (downward), while the harness works opposite to the fall (upward).

(23)   \begin{equation*} mgd_s cos(0) + F_h d_s cos(180^{\circ})= -mgd \end{equation*}

Calculating the cos(0) = 1 and cos(180^{\circ}) = -1

(24)   \begin{equation*} mgd_s  - F_h d_s = -mgd \end{equation*}

Adding F_h d_s to the both sides and mgd to both sides:

(25)   \begin{equation*} mgd_s  + mgd = F_h d_s \end{equation*}

Dividing everything by the stopping distance d_s:

(26)   \begin{equation*} mg  + \frac{mgd}{d_s} = F_h \end{equation*}

We see that the average force applied while stopping a fall is the weight plus the weight multiplied by the ratio of the fall distance to stopping distance. A longer stopping distance will decrease the average force applied.

We can factor out the weight to simplify:

(27)   \begin{equation*} mg (1  + \frac{d}{d_s}) = F_h \end{equation*}

From this formula we can see that the average stopping force will always be (1  + \frac{d}{d_s}) times greater than the weight. For a stopping distance equal to the fall distance, the force will be 2x the weight. For fall distance twice the stopping distance the average stopping force will be 3x the weight.

Our stopping distance was 2 m and our fall distance was 8 m, so we expect an average stopping force 5x greater than the weight.

(28)   \begin{equation*} (65\,\bold{kg})(9.8\,\bold{m/s/s})(1+ \frac{8\, \bold{m} }{2\, \bold{m}}= F_h \end{equation*}

The final result:

(29)   \begin{equation*} F_h = 3,185\, bold{N} \end{equation*}

When our stopping distance was only 2 cm,  or 0.02 m, the falling distance was 400x greater than the stopping distance. We expect an average stopping force that is 401x greater than the weight!

(30)   \begin{equation*} (65\,\bold{kg})(9.8\,\bold{m/s/s})(1+ \frac{8\, \bold{m} }{0.02\, \bold{m}}= F_h \end{equation*}

The final result:

(31)   \begin{equation*} F_h = 25,5437\, \bold{N} \end{equation*}

Notice that in solving this problem we found a general formula for the average force required to stop a fall (F_h = mg (1  + \frac{d}{d_s})), but we should keep in mind that this formula only provides us with the average force. The actual peak force during the impact is greater than the average force. The peak force can be determined by modeling how the rope stretches and how much kinetic energy is transferred to thermal energy in the rope. To analyze how this dissipation of energy occurs within the rope, as well as in tendons and bones, we need to include those objects in our system (rather than just the climber) and learn about and .  We will do that soon.

Reinforcement Exercises

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

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