# Studying Help

1. Jun 15, 2006

### JayDub

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

(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.

2. Jun 15, 2006

### illwerral

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.

3. Jun 15, 2006

### JayDub

Well, what about in the case such as:

A 5kg block initially travelling 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?

4. Jun 15, 2006

### Pyrrhus

You can use

$$\Delta E_{mechanical} = W_{friction}$$

5. Jun 15, 2006

### JayDub

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...

6. Jun 15, 2006

### Pyrrhus

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.