How Much Work is Done by the Cake Batter on a Falling Worker?

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Homework Help Overview

The problem involves a 65-kg worker who falls into a vat of cake batter after losing his balance, traveling a total distance of 6.0 m before coming to rest. The discussion centers around calculating the work done on the worker by the cake batter, utilizing concepts from physics such as gravitational potential energy and kinetic energy.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the worker's fall and the forces acting on him, including gravitational force and the work done by the cake batter. There are attempts to calculate work using the displacement and gravitational potential energy, while some participants question whether all energy lost during the fall has been accounted for.

Discussion Status

There is an ongoing exploration of the calculations involved in determining the work done by the cake batter. Some participants have provided guidance on the relationship between gravitational potential energy and the work done, while others are seeking clarification on whether additional factors need to be considered in their calculations.

Contextual Notes

Participants are navigating the implications of energy loss during the fall and through the batter, with attention to significant figures in their calculations. The discussion reflects uncertainty regarding the completeness of their approaches and the assumptions made about energy transfer.

Jamest39
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Homework Statement


A 65-kg worker at a bakery loses his balance and falls 4.0 m before hitting the surface of a large vat of cake batter. He continues to travel downwards an additional 2.0 m before the cake batter finally brings him to rest. Calculate the work done on the worker by the cake batter.

Homework Equations


F = ma
ΔKE = Wnet
W = Fscosθ
PEgravity = mgh

The Attempt at a Solution


ΔKE = KEfinal - KEinitial = 0 - KEinitial ⇒ KEintitial = -(1/2)mv^2
PEgravity = mgh = (65 kg)(9.80 m/s^2)(4.0 m) = 2548 J
 
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Consider that the worker is still falling (moving vertically) through the cake batter until he comes to rest.
 
gneill said:
Consider that the worker is still falling (moving vertically) through the cake batter until he comes to rest.

So,
F = ma = (65 kg)(9.80 m/s^2) = 637 N
W = Fscosθ = (637 N)(6.0 m)cos180° = -3822 J (6.0 as the displacement considering his whole movement?)
 
That will work. The basic premise is that the work falls a total of 6 m through the gravitational field and so gains energy Mgh from the change in gravitational potential energy. That total amount of energy is "lost" to the cake batter over distance that he travels through the batter.
 
gneill said:
That will work. The basic premise is that the work falls a total of 6 m through the gravitational field and so gains energy Mgh from the change in gravitational potential energy. That total amount of energy is "lost" to the cake batter over distance that he travels through the batter.

Was my answer right or do I still have to factor in the energy he lost while going through the batter?
 
Jamest39 said:
Was my answer right or do I still have to factor in the energy he lost while going through the batter?
Your answer was fine. The energy lost to the batter is the energy gained from falling. That's why the worker comes to rest in the batter. That energy represents the work done by the batter. The only way you could improve your result would be to make sure that its stated to the correct number of significant figures.
 

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