Confusion using a simple formula Physics EK 1001# 189

[BeatMCAT]

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So, I just got stumped doing a very simple problem. Its from EK 1001.
If theta is 30˚, d is 10m and x is 10m, how long after mass reaches the bottom of plane does it req to move distance x?

d is the hypotenuse of the inclined plane, and x is distance on flat ground at the end of inclined plane.

I was able to figure out that the a=5m/s2 on the incline. Therefore, the velocity at the end of incline is 10 m/s. Then, I decided to fiure out time using D=V(avg) * time. So i got Vavg to be 5 m/s and therefore t=2 seconds. HOWEVER, the answer is 1 second, which means they did not use Vavg. So my question is WHEN to use this formula to find avg and when not to use it. Thanks a lot!

 

[vayntraubinator]

I was able to figure out that the a=5m/s2 on the incline. Therefore, the velocity at the end of incline is 10 m/s.

Think about that. You cant get final velocity from acceleration if you didnt know the time..

m/s^2 * (time(s)) = m/s

 

[bullsfan1986]

So initially you set 1/2mv^2=mgh and solve for v=10m/s (I think that you got that).

Now that you have velocity you can set work= KEf-KEi. Because your KEi = 0, KEf= W. Because W= ma x d, you can set ma x d = 1/2mv^2 and plug in your velocity of 10m/s. Because m cancels out and d=10m, your a=5m/s^2 (I think you got that part too).

This just means that the block is travelling at 10m/s with an acceleration of 5m/s^2 when it is at the bottom of the incline..but the problem is asking for you to figure out how much time it takes to get to the end of x.

The thing that I did after this was just to set ma= m (v)/t where mass will cancel out, so you will have a= (v)/t. Because you know that a= d/t^2 just plug that in for a so that d/t^2=(v)/t. Set v=10m/s and solve for t.

 

[BeatMCAT]

Thanks a lot! that makes sense.

 

[inaccensa]

So initially you set 1/2mv^2=mgh and solve for v=10m/s (I think that you got that).

Now that you have velocity you can set work= KEf-KEi. Because your KEi = 0, KEf= W. Because W= ma x d, you can set ma x d = 1/2mv^2 and plug in your velocity of 10m/s. Because m cancels out and d=10m, your a=5m/s^2 (I think you got that part too).

This just means that the block is travelling at 10m/s with an acceleration of 5m/s^2 when it is at the bottom of the incline..but the problem is asking for you to figure out how much time it takes to get to the end of x.

The thing that I did after this was just to set ma= m (v)/t where mass will cancel out, so you will have a= (v)/t. Because you know that a= d/t^2 just plug that in for a so that d/t^2=(v)/t. Set v=10m/s and solve for t.

I got the Vf= 10m/s. This is the velocity at the bottom, so my next step was to use vf2=v2+2ax. The value of x is 10m, and vi in this case is 10m/s, so solving the equation gives me a=5m/s. Finally, I used vf=v+at, again vf=0, t=2. cyan anyone explain what is wrong with what i did

 

[dorjiako]

I got the Vf= 10m/s. This is the velocity at the bottom, so my next step was to use vf2=v2+2ax. The value of x is 10m, and vi in this case is 10m/s, so solving the equation gives me a=5m/s. Finally, I used vf=v+at, again vf=0, t=2. cyan anyone explain what is wrong with what i did

You got a = 5m/s^2, The velocity on using the incline plane buttom is V = (2ad)^1/2. V = 10m/s. use this velocity to find time. It is okay to use this velocity to find the time since INITIAL HORIZONTAL VELOCITY IS EQUAL TO FINAL HORIZONTAL VELOCITY. x/v = 10/10 = 1s.
You can also use the formular v^2= 2ax therefore (x^2/t^2) = 2ax , 100/t^2= 2x5x10,
100/t^2= 100
t^2=1 therefore t=1.