Max Height of 6 kg Object with Air Resistance

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Homework Help Overview

The discussion revolves around determining the maximum height reached by a 6 kg object projected upwards with an initial speed of 15 m/s, while experiencing air resistance. Participants are exploring the effects of forces acting on the object, including gravitational force and air resistance, within the context of kinematics and energy conservation principles.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to apply energy conservation principles and kinematic equations but struggles with the inclusion of air resistance. Some participants question the treatment of forces and acceleration, particularly how to account for air resistance in the calculations.

Discussion Status

Participants are actively engaging with the problem, offering insights into the direction of forces and the implications for acceleration. There is a recognition of the complexity introduced by air resistance, and some guidance has been provided regarding the signs of forces and accelerations.

Contextual Notes

There is an ongoing exploration of different scenarios involving forces, including a separate example of a block moving up an incline with friction. The discussions highlight the need for clarity on how to apply energy principles and account for various forces in different contexts.

JayDub
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Hey there, I have been studying for my physics final coming up and I am constantly getting stuck on the stuff that we started at the beginning of the year. Such as this question:

A 6 kg object is projected directly upwards with an initial speed of 15 m/s. This object experiences an average air resistance force of 24 N. What is the maximum height reached by this object?

I have no idea how to solve it. If it weren't for the air resistance I would prolly go something along the lines of:

KE = PE
(1/2)mv^2 = mgh
h = [(1/2)v^2] / g

I also thought of going like

Fnet = ma
24 N = (6 kg)(a)
a = 4 m/s

So the total acceleration would be: -9.8 m/s/s + 4 m/s/s which is -5.8 m/s/s

Then I used kinematics such as

Vf^2 = Vo^2 + 2ad
(0m/s)^2 = (15 m/s)^2 + 2(-5.8m/s/s)d

d = 19.3 m

That is one of the answers but it is wrong as the answer is 8.2m. So I have no idea how to solve questions like these, so I am coming here to ask for help. Thank you.
 
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JayDub said:
So the total acceleration would be: -9.8 m/s/s + 4 m/s/s which is -5.8 m/s/s

Your problem is that air resistance is in the direction opposite to the velocity of the object, ie down. Since gravity is also acting in the down direction the two forces have the same sign...so if gravity is negative, so's the acceleration due to wind resistance.
 
Well, what about in the case such as:

A 5kg block initially traveling at 11 m/s up a 30 degree incline. A frictional force of 9.4N acts on the block as it moves up the incline. What is the maximum height the block will reach?
 
JayDub said:
Well, what about in the case such as:

A 5kg block initially traveling at 11 m/s up a 30 degree incline. A frictional force of 9.4N acts on the block as it moves up the incline. What is the maximum height the block will reach?

You can use

[tex]\Delta E_{mechanical} = W_{friction}[/tex]
 
E mechanical = W friction?

I do not quite see how that will help me...

PE = KE = Fd?

mgh = .5(m)v^2 = 9.4N(d)

I do not have a distance...
 
I meant

[tex]E - E_{o} = W_{friction}[/tex]

With the use of some trigonometry and analyzing the initial energy and the final energy conditions, you can set up the equation properly.
 

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