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

AI Thread Summary
The discussion focuses on calculating the work done on a 65-kg worker by cake batter after he falls 4.0 m and continues 2.0 m further before coming to rest. The gravitational potential energy lost during the fall is calculated as 2548 J, while the work done by the batter is determined to be -3822 J, accounting for the total distance fallen. The energy lost to the batter is equal to the energy gained from the fall, which explains why the worker stops moving in the batter. The calculation is confirmed to be correct, with a suggestion to ensure significant figures are properly noted. Overall, the work done by the cake batter is directly related to the energy lost by the worker during the fall.
Jamest39
Messages
34
Reaction score
1

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
 
Physics news on Phys.org
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.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top