What is the net force on Hydro in Problem #7 of the Net Force Homework?

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Homework Help Overview

The discussion revolves around a physics problem involving a character named Hydro, who falls a distance of 4.67 m and comes to rest after 0.543 seconds upon hitting the water. Participants are exploring the net force exerted on Hydro during this time, which falls under the subject area of dynamics and kinematics.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants are attempting to use impulse and momentum concepts to analyze the problem. Some are questioning how to calculate the speed of Hydro upon impact and how to derive acceleration from the given data. There are discussions about the velocity vs. time graph and the interpretation of negative acceleration.

Discussion Status

Several participants have offered guidance on breaking the problem into two parts: determining Hydro's speed upon hitting the water and calculating the net force during deceleration. There is an ongoing exploration of the basic formulas and concepts involved, with some expressing confusion and seeking clarification.

Contextual Notes

Participants are grappling with the application of physics concepts, particularly regarding the transition from free fall to deceleration upon hitting the water. There is a noted struggle with understanding the implications of the velocity vs. time graph and the calculations involved.

jg871
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Homework Statement


Problem #7[/B](Due 9/26 Friday by 8:00 a.m.)
Hydro, who has a mass of 45.1 kg, falls straight down 4.67 m to the water below.
a. If Hydro comes to rest .543 seconds after hitting the water, what was the net force exerted on Hydro over this time?

Homework Equations


The Attempt at a Solution


Looked in the book, read the sections the question pertains to, and am still lost. Any assistance with how to solve this would be appreciated.[/B]
 
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You can use impulse to solve this.
 
(4.67/.543) *45.1?
 
jg871 said:
(4.67/.543) *45.1?
No.

What's the momentum of Hydro as he first touches the water?
 
Doc Al said:
No.

What's the momentum of Hydro as he first touches the water?
4.67/.543 so 8.60?
 
jg871 said:
4.67/.543 so 8.60?
M/s
 
jg871 said:
4.67/.543 so 8.60?
Don't just give numbers--show how you used the basic formulas.

In any case, no. Answer this: If something falls a distance X, how fast is it going at that point?
 
Doc Al said:
Don't just give numbers--show how you used the basic formulas.

In any case, no. Answer this: If something falls a distance X, how fast is it going at that point?
Acceleration of gravity correct?
 
jg871 said:
Acceleration of gravity correct?
The acceleration of a falling object is g, the acceleration due to gravity. But what's the speed of the object after it falls?
 
  • #10
Doc Al said:
The acceleration of a falling object is g, the acceleration due to gravity. But what's the speed of the object after it falls?
I'm struggling mightily.
 
  • #11
jg871 said:
I'm struggling mightily.
jg871 said:
I'm struggling mightily.
jg871 said:
I'm struggling mightily.
What may seem obvious is just not registering for me here
 
  • #12
Doc Al said:
Don't just give numbers--show how you used the basic formulas.

In any case, no. Answer this: If something falls a distance X, how fast is it going at that point?

Doc Al said:
The acceleration of a falling object is g, the acceleration due to gravity. But what's the speed of the object after it falls?

speed = distance / time
so
4.67m / 0.543s
but how do you get the acceleration from that?
 
Last edited:
  • #13
jg871 said:

Homework Statement


Problem #7[/B](Due 9/26 Friday by 8:00 a.m.)
Hydro, who has a mass of 45.1 kg, falls straight down 4.67 m to the water below.
a. If Hydro comes to rest .543 seconds after hitting the water, what was the net force exerted on Hydro over this time?

Homework Equations


The Attempt at a Solution


Looked in the book, read the sections the question pertains to, and am still lost. Any assistance with how to solve this would be appreciated.[/B]
I'm also working on this question for homework.
For the velocity vs. time graph, does your constant negative line start at some velocity above zero and stop at zero? Or does it go past zero into a negative, or does it start at zero and go down?
Wasn't very good at drawing these for the test either...
 
  • #14
Ering said:
speed = distance / time
so
4.67m / 0.543s
but how do you get the acceleration from that?
Divide by time again
 
  • #15
Ering said:
I'm also working on this question for homework.
For the velocity vs. time graph, does your constant negative line start at some velocity above zero and stop at zero? Or does it go past zero into a negative, or does it start at zero and go down?
Wasn't very good at drawing these for the test either...
I wasn't either. Go to umd? I believe it goes into a negative but I'm confused with the class so far
 
  • #16
Ering said:
speed = distance / time
That will give you average speed.

Ering said:
so
4.67m / 0.543s
You are dividing the distance it falls (before it hits the water) by the time it takes to come to rest (after it hits the water). That won't help.

Separate this problem into two parts: (1) How fast is it moving when it hits the water? (2) What net force does it have after it hits the water and is slowing down?

Solve each part separately.

Ering said:
but how do you get the acceleration from that?
The acceleration due to gravity is a constant. Look it up! (It's usually called g.)
 

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