# Block of unknown mass projected up incline.

1. Sep 30, 2009

### amb1989

A block is projected up a frictionless inclined plane with initial speed of 3.50 m/s. The angle of incline is 32 degrees. How far up the planes does the block go? How long does it take to get there? What is its speed when it gets back to the bottom.

$$\vec{F}_{net} = \Sigma \vec{F} = m \vec{a}$$

$$(\vec{F}_{net})_x = \Sigma F_x = 0$$
$$(\vec{F}_{net})_y = \Sigma F_y = 0$$

I don't really understand how to start this problem. The only thing I could think of was to break the velocity into x and y components and once I do that I really don't know where to go from there...

(my first post so I'm sorry if I did not follow the guidelines of everything should be presented)

2. Sep 30, 2009

### w3390

The velocity the give you is in the direction of the plane. What you need to do is draw a picture and make a free-body diagram. You will see that the only force acting on the block is a force trying to pull it back down the ramp. You will be able to solve this equation for acceleration. You can then use this acceleration and the velocity given to find the distance it travels and how long it takes for the block to stop. For the last part, it is a simple kinematic equation using the force already found to figure out its speed at the bottom of the ramp.

3. Sep 30, 2009

### tiny-tim

Welcome to PF!

Hi amb1989! Welcome to PF!
You can answer these two by using conservation of energy.
You have s from the first answer. You also have u and v from the question.

So use two of the standard constant acceleration equations to find a, and then t.

4. Sep 30, 2009

### amb1989

I drew out the free body diagram before posting this topic and I see that force trying to pull the block down the ramp. The thing is how am I supposed to solve for the acceleration when I don't know the mass of the block or the intial force? I feel like I don't have enough information to do this problem.

I also tried the conseravation of energy method and I couldn't get it that way either. Here's what I got when I tried that:

(1/2)mv^2 = mgh

I cancel mass and divide by g

((1/2)v^2)/g = h

plugging in I get:
((1/2)(3.5)^2)/9.81 = h

The answer that I got for that is 0.625m and it should be 1.18m. The reason I didn't mention that I did this in the first place is that we haven't covered conservation of energy yet and I'm even sure if I did that right...Well obviously I didn't do that right seeing as I didn't get the right answear.

Last edited: Sep 30, 2009
5. Sep 30, 2009

### w3390

The reason why it seems like you don't have enough information is because the mass is supposed to cancel when you write your equation. You should notice that you have mass on the left and right side. They just cancel. Also, think of it as looking at the block beginning after some force has already been applied to it to make it move initially at 3.5 m/s. By the time you are examining it, the only force is the one pulling it back down the ramp, so there is no force in the direction up the ramp.

6. Sep 30, 2009

### amb1989

Thanks for taking the time to reply to my problem. However I've already established what you said in your post. In my post before yours I went through exactly what I did and that included the free body diagram that I drew and what I did with the conservation of energy equation (yes I did cancel mass ;). It just doesn't seem to work out when I try that and I'm not sure what I'm doing wrong with it.

7. Sep 30, 2009

### w3390

Okay, post what your equation looks like and put the number in step by step. I'll compare it to what I did and see what's wrong.

8. Sep 30, 2009

### amb1989

(1/2)mv^2 = mgh

I cancel mass and divide by g

((1/2)v^2)/g = h

plugging in I get:
((1/2)(3.5)^2)/9.81 = h

The answer that I got for that is 0.625m and it should be 1.18m.

9. Sep 30, 2009

### tiny-tim

Hint: .0625 = 1.18 times sin32º

10. Sep 30, 2009

### amb1989

oh my god...thank you haha I never would have made the connection. thanks a lot!

11. Sep 30, 2009

### tiny-tim

he he

ok … can you find a and t now?

(and I'm gong to bed :zzz: …)​

12. Sep 30, 2009

### amb1989

Ok I figured everything out, thanks a lot everyone!