Max Height of 6 kg Object with Air Resistance

AI Thread Summary
A 6 kg object projected upwards at 15 m/s experiences 24 N of air resistance, complicating the calculation of its maximum height. The initial approach using kinetic and potential energy equations is flawed due to the need to account for air resistance as a downward force. Correctly, the net acceleration should consider both gravity and air resistance, leading to a total acceleration of -5.8 m/s². This adjustment results in a maximum height of 8.2 m, which is the accurate solution. Understanding the influence of opposing forces like air resistance is crucial for solving such physics problems effectively.
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 beggining 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

\Delta E_{mechanical} = W_{friction}
 
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

E - E_{o} = W_{friction}

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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