Work Done and Energy Output (Walking vs Running)
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Description
This is a basic physics GCSE tutorial exploring in simple terms how work done and energy output varies when walking or running a set distance. Video tutorial to come.
The 'problem'
How much work does a person of mass 50kg do if they run or walk 5km in 30 minutes and 60 minutes respectively?
Some 'assumptions'
There are four assumptions we're going to make in order to simplify this problem:
- The person starts and ends at rest in both cases;
- Once moving at the necessary velocity, the person travels at a constant velocity in both cases;
- Both 5 km routes are exactly the same, on a flat track with no hills, descents etc.; and
- Lastly, we assume that it takes the same amount of time to reach average velocity (accelerate) and to come to a stop (decelerate) in both cases; we will assume 2 seconds in this problem for both acceleration and deceleration, but it really doesn't matter what value you choose.
To simplify the calculation, we're also going to ignore friction and air resistance, as well as any potential vertical bounce associated with running.
There are two methods
There are two ways of looking at this. The first is to calculate the work done in each scenario; the second is to look at how much kinetic energy is gained and lost. Each method should give us the same answer.
How do we know this? Because one way of defining kinetic energy is : the amount of work needed to accelerate / decelerate an object.
Method 1: Work done
To solve this problem, all we need to do is calculate how much work is required to
a) get a person moving at the two different average velocities from rest, and
b) to come to a stop from those two velocities.
What about the time in between though? Well, based on the above assumptions, the rest of the time in between (i.e. most of the walking/running) does not require any additional force. This is because the person is traveling at a constant velocity during this time - and remember, additional forces are only required if there is a change in velocity. As no additional forces are required, no additional work is done. Below are some equations that we will be using for our calculations.
Useful Equations
- Velocity = displacement ÷ time
- Acceleration = (velocity_{final} - velocity_{initial}) ÷ time
- Force = mass x acceleration
- Work = force x distance
- Distance = (velocity_{initial} x time) + (^{1}/_{2} x acceleration x time^{2}) <-- Look out for relevant tutorial
Work done walking 5km | Work done running 5km |
Step 1: Calculate average velocity | |
Velocity = displacement ÷ time | |
Velocity = 5000 m ÷ 3600 s | Velocity = 5000 m ÷ 1800 s |
Velocity = 1.39 m/s | Velocity = 2.78 m/s |
Step 2: Calculate rate of acceleration | |
Acceleration = (velocity_{final} - velocity_{initial}) ÷ time | |
Acceleration = (1.39 m/s - 0 m/s) ÷ 2 seconds | Acceleration = (2.78 m/s - 0 m/s) ÷ 2 seconds |
Acceleration = 0.695 m/s^{2} | Acceleration = 1.39 m/s^{2} |
Step 3: Calculate force needed to accelerate | |
Force = mass x acceleration | |
Force = 50 kg x 0.695 m/s^{2} | Force = 50 kg x 1.39 m/s^{2} |
Force = 34.8 N | Force = 69.5 N |
Step 4: Calculate distance over which force was acting | |
Distance = (velocity_{initial} x time) + (^{1}/_{2} x acceleration x time^{2}) | |
Distance = (0 m/s x 2 s) + (^{1}/_{2} x 0.695 m/s^{2} x (2 s^{2})) | Distance = (0 m/s x 2 s) + (^{1}/_{2} x 1.39 m/s^{2} x (2 s^{2})) |
Distance = 0 m + (^{1}/_{2} x 0.695 m/s^{2} x 4 s^{2}) | Distance = 0 m + (^{1}/_{2} x 1.39 m/s^{2} x (2 s^{2})) |
Distance = 1.39 m | Distance = 2.78 m |
Step 5: Calculate work done to get to average velocity | |
Work done = force x distance | |
Work done = 34.8 N x 1.39 m | Work done = 69.5 N x 2.78 m |
Work done = 48 J | Work done = 193 J |
Step 6: Calculate work done to decelerate to a stop |
This will be the same as the work done to accelerate to average velocity - the force is simply being applied in the opposite direction. As work is a scalar measurement the answer is exactly the same as that calculated in Step 5.
Work done = 48 J | Work done = 193 J |
Step 7: Calculate total work done | |
Total work done = Work_{acceleration} + Work_{deceleration} | |
Total work done = 48 J + 48 J | Total work done = 193 J + 193 J |
Total work done = 96 J | Total work done = 386 J |
Method 2: Gain/Loss of Kinetic Energy
We haven't delved into kinetic energy in much detail, however you may have done so in school or elsewhere already. If that's the case, follow along with this method. Below is the additional equations we need for this method; refer to step 1 in Method 1 above to work out the average velocity for each journey.
Useful Equations
- Kinetic energy _{object at any given velocity}= ^{1}/_{2} x mass x velocity^{2}
- Kinetic energy _{change in velocity}= ^{1}/_{2} x mass x (velocity_{initial} - velocity_{final})^{2}
Kinetic energy for walking 5km | Kinetic energy for running 5km |
Step 1: Calculate kinetic energy to accelerate | |
Kinetic energy = ^{1}/_{2} x mass x (velocity_{initial} - velocity_{final})^{2} | |
Kinetic energy = ^{1}/_{2} x 50 kg x (1.39 m/s - 0 m/s)^{2} | Kinetic energy = ^{1}/_{2} x 50 kg x (2.78 m/s - 0 m/s)^{2} |
Kinetic energy = ^{1}/_{2} x 50 kg x 1.93 | Kinetic energy = ^{1}/_{2} x 50 kg x 7.73 |
Kinetic energy = 48 J | Kinetic energy = 193 J |
Step 2: Calculate kinetic energy to decelerate to a stop | |
Kinetic energy = ^{1}/_{2} x mass x (velocity_{initial} - velocity_{final})^{2} | |
Kinetic energy = ^{1}/_{2} x 50 kg x (0 m/s - 1.39 m/s)^{2} | Kinetic energy = ^{1}/_{2} x 50 kg x (0 m/s - 2.78 m/s)^{2} |
Kinetic energy = ^{1}/_{2} x 50 kg x 1.93 | Kinetic energy = ^{1}/_{2} x 50 kg x 7.73 |
Kinetic energy = 48 J | Kinetic energy = 193 J |
Step 3: Calculate total kinetic energy used | |
Total kinetic energy = KE_{acceleration} + KE_{deceleration} | |
Total kinetic energy = 48 J + 48 J | Total kinetic energy = 193 J + 193 J |
Total kinetic energy = 96 J | Total kinetic energy = 386 J |
As you can see above, the kinetic energy expended is the same as the work done by the individual to walk and run 5 km in 60 and 30 minutes respectively.