#1

If the length of a ramp is 2.26m, and it takes 3.12s for a marble to roll down the ramp, what is the angle of the ramp?

There is no friction in this problem.

I'm trying to figure this out but I am in epic fail. How would I solve this?

Btw, this was a lab, so I know the angle is less than 45, and it's probably around 20, but I don't understand how to calculate for it.

There is no friction in this problem.

I'm trying to figure this out but I am in epic fail. How would I solve this?

Btw, this was a lab, so I know the angle is less than 45, and it's probably around 20, but I don't understand how to calculate for it.

#2

there's an elephant in, the way !!!!

#3

There's an elephant in the way!

Edit: shit

Edit: shit

#4

First figure out the acceleration of the block by dividing the distance in meters by time in seconds

#5

#6

First figure out the acceleration of the block by dividing the distance in meters by time in seconds

that would give you the velocity i believe.

maybe you could solve the for the velocity in the way mentioned above and then use a kinematics equation involving distance and velocity. The horizontal distance should be set to (length of ramp)cos(angle of ramp). The problem is that there aren't any equations that don't also involve acceleration, and i can't think of a way to solve for the acceleration.

try the math thread.

#7

Once you have that acceleration, use the equality

gsin(theta) = a

Where g is gravity(9.8 m/s^2), and a is the acceleration you just found.

Then you just solve out for theta

EDIT: Above is right! to find acceleration you take the root of the time and divide it into of two times the distance!

gsin(theta) = a

Where g is gravity(9.8 m/s^2), and a is the acceleration you just found.

Then you just solve out for theta

EDIT: Above is right! to find acceleration you take the root of the time and divide it into of two times the distance!

*Last edited by Pinto111 at Aug 18, 2009,*

#8

First figure out the acceleration of the block by dividing the distance in meters by time in seconds

I do that to get 0.724m/s. Then I divide that by the time again to get a=0.232m/s squared. Since on a ramp the acceleration=(gravity)(sin of angle), i use the equation Angle=Invsin(acceleration/gravity), and i get 1.35 degrees, which is obviously wrong.

What am I doing wrong?

#9

Acceleration = 2.6 m/s^2

#10

Acceleration = 2.6 m/s^2

? how did you get that?

That doesn't seem to make sense though, because it's saying it takes a second for it to go 2.6 meters, when it took more time to travel a smaller distance.

AP Physics is effing hard

#11

Angle = 15.6 deg with that accel

#12

? how did you get that?

That doesn't seem to make sense though, because it's saying it takes a second for it to go 2.6 meters, when it took more time to travel a smaller distance.

AP Physics is effing hard

You use the eqution

x = V1t + .5at^2

where x is distance, V1 is initial velocity(here is zero), a is acceleration and t is time

#13

You use the eqution

x = V1t + .5at^2

where x is distance, V1 is initial velocity(here is zero), a is acceleration and t is time

Ohhh so I can't just simply divide the velocity by time again to get the acceleration?

#14

Ohhh so I can't just simply divide the velocity by time again to get the acceleration?

Nope, I misspoke in my fist post.

Because the velocity is changing throughout the balls journey, when you divide distance by time you actually get the

__average__velocity

So a different equation is needed to get the acceleration

#15

Nope, I misspoke in my fist post.

Because the velocity is changing throughout the balls journey, when you divide distance by time you actually get theaveragevelocity

So a different equation is needed to get the acceleration

So in the equation you posted, x isn't really the horizontal distance but just the distance of the ramp?

Assuming that, I would plug in as 2.26=(0.5)a(3.12)^2

And I would get a=0.462m/s^2.

How do did you get the acceleration you got? Because when I plugged that into the equation for the angle, I got a reasonable angle, but when I plugged this accelration in I ended up with about 3 degrees

#16

hmm seems i f

I actually took the root of the time for some reason..... hmmmmm

*u*cked up...I actually took the root of the time for some reason..... hmmmmm

#17

Why is physics so hard

#18

Usually I'm good aat it but I havent done any in months lol

#19

Just a guess, but wouldn't it just be something like Sin(Theta) * 9.8 = 3.12, giving you 19.05 degrees?

Edit: Nevermind, wrong. :p

Edit: Nevermind, wrong. :p

#20

s=ut + 0.5at^2

a = g cos theta

That should help

a = g cos theta

That should help

#21

forbidden thread.