A 50 kg window washer of skyscrapers falls 30 m from the top of a small building and lands stiff-legged in some mud next to the safety net where he was supposed to land. The man stops after sinking...
A 50 kg window washer of skyscrapers falls 30 m from the top of a small building and lands stiff-legged in some mud next to the safety net where he was supposed to land. The man stops after sinking 0.80 m into the mud. Find the average froce of the mud on the person while bringing him to rest.
In order to answer this question, we need to recognize the nature of what a force is. Newton's 2nd Law tells us that:
Force = Mass x Acceleration
We know that the window washer hits the mud with some amount of velocity, and the mud will slow him down (accelerate him) to a final speed of 0. We need to find the value of that acceleration.
Acceleration is equal to a change in velocity over time. We know the man's mass (50kg) but we don't know the acceleration that the mud provides, or the time that he spends sinking in it. We could deduce the acceleration by finding the velocity with which the man strikes the mud, but this would take several kinematic equations to solve. Instead, we'll use energy.
Recall that Potential Energy and Kinetic Energy are related through the Conservation of Energy: `PE_i + KE_i = PE_f + KE_f`
We also know that `PE = mgh` and `KE = .5mv^2`
We need to treat this problem in three steps.
- Step one will address the window washer's fall from a 30m height, until he contacts the mud.
- Step two will address the window washer sinking into the mud.
- In the final step, we will solve for the force of the mud.
Step 1: Falling
PE = mgh
Fortunately we already know the values of m, g and h;
- m = 50kg
- g = 9.8m/s^2
- h = 30m
`PE = (50)(9.8)(30) `
`PE = 14700 J`
When the window washer falls the 30m height, he converts all of his potential energy into kinetic energy. This gives him velocity.
KE = 14700 Joules
Step 2: Sinking
When the window washer sinks into the mud, his final velocity is 0. All of his energy is "burned away" as friction; his final PE and KE are both 0.
Step 3: Solving for F
Recall that both energy and work are measured in forces acting over a distance. In this case, the mud is exerting a force on the washer over a distance of 0.80m which will bring him to a stop (i.e. completely equal the kinetic energy he possessed at the moment he contacted the mud.
`F = ma and W = Fd`
`KE = mad`
`14700J = (50kg)(a)(0.80m)`
kg and m cancel out, leaving us with an answer in `m/s^2` , acceleration
`367.5 = a`
`F = ma`
`F = (50)(367.5)`
`F = 18375`
since we're limited by two significant figures, `F = 18000N`
If this seems like a ridiculous answer, it is. Using `KE = .5mv^2` , we would see that the washer strikes the mud at a speed of 24 meters per second, or 54 miles per hour. It's going to take a lot of force to bring him to a stop in less than one meter!