Calculate the horizontal speed of the box as it falls

  • Thread starter paigegail
  • Start date
In summary, Paige provided a summary of the conversation and explained that she got the answer to the puzzler using the method she described to the other participant. Then, she provided an equation for calculating the acceleration, which she got wrong. She thanks the other participant for being patient with her and ends the conversation.
  • #1
paigegail
22
0
We were given the answer to this puzzler and its 71.96875 N
 
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  • #2
Yes, that's the same answer I got using the method I described to you.
 
  • #3
What I did...

I will show you what I did and then can you please tell me what I'm doing wrong.

Consider Vertical:
¥Äx= -12.96 m
Vo= 0 m/s
a= -9.80 m/s
v= ?
t =?

v©÷ = vo©÷ + 2a¥Äx
v©÷ = 0 + 2a¥Äx
= 2a¥Äx
= 2(-9.80)(-12.96)
= 254.016
= ¡î254.016
=15.93 m/s

v = vo + at
15.93 = 0 + (-9.80)t
t= 1.626

Consider horizontal
¥Äx = 7.5 + 5= 12.5 m
t = 1.626 s
v= ?
¥Äx = vt
v= ¥Äx/t
= 7.688 m/s

What do I do from here and is this really right?
 
  • #4
The equations in the last one are a little wack. sorry...
© = squared
¥Äx= delta x
 
Last edited:
  • #5


Originally posted by paigegail

Consider horizontal
¥Äx = 7.5 + 5= 12.5 m
You calculated the time to fall correctly. But you are using the wrong horizontal distance to calculate the horizontal speed of the box as it falls:

distance = 7.5m (do not add the distance the box was pushed!)
 
  • #6
Ok...
so I got vx to be 4.613 m/s. Then I use Pythagoras to find the velocity (vy-squared- + vx-squared- = v-squared) and I got it to be 16.59. Then to find acceleration I went:
a= V/t
= 10.20 m/s-squared-

Am I going right...because then I get F= ma
F= 10.8 (10.20)
=110.19 N

I don't think that's right...


By the way...Thanks for all of your help!

Oh and do you have a messenger system of some sort because that would make this much easier.
 
  • #7
At the moment the box leaves the cliff, Vy = 0. The only speed is horizontal, which you just calculated: Vx.

So now attack the first part of the motion: the box being pushed along the ground. Tell us everything you know about that segment of the motion, from initial push until the box sails off the cliff.
 
  • #8
Delta X = 5.0 m
Vx= 4.613 m/s
t =?

Delta X = Vx(t)
t= delta x/ vx
= 5/4.613
= 1.083 s

So do I add 1.083 to the previously establish 1.626 and then get 2.709 for total time?
 
  • #9
You don't care about the total time; you want the acceleration. You know the initial velocity v0=0 m/s, the final velocity v=4.61 m/s, and the distance x=5 m. There is a kinematics formula that relates those three quantities to the acceleration a, without involving time.
 
  • #10
Ok...so I have that. But what is SOOOOO EXTREMELY FRUSTRATING...IS WHAT DO I DO AFTER I HAVE ALL THAT! The acceleration becomes 0.4163 m/s-squared-. F= ma... that makes is 10.8(0.4163)=4.49604 N that's no where near the answer.
 
  • #11
Originally posted by paigegail
Ok...so I have that. But what is SOOOOO EXTREMELY FRUSTRATING...IS WHAT DO I DO AFTER I HAVE ALL THAT! The acceleration becomes 0.4163 m/s-squared-. F= ma... that makes is 10.8(0.4163)=4.49604 N that's no where near the answer.
Relax...

First step is to calculate the acceleration properly. I don't think you did. Tell us how you did it.

Once you find the acceleration, then apply F=ma properly. What are all the forces acting on the box?
 
  • #12
K...I found acceleration wrong. I realized that.
Acceleration = 1.733 m/s -squared-

So then do I go F=ma
F= 10.8 (1.733)
= 18.7164

So is that Fnet? If it is...then what?
 
  • #13
Originally posted by paigegail
K...I found acceleration wrong. I realized that.
Acceleration = 1.733 m/s -squared-
How did you calculate the acceleration?
 
  • #14
v-squared- = vo-squared- + 2aDelta X
4.163 -squared- = 0 + 2(5)a
17.330569 =10a
a=1.733
 
  • #15
Originally posted by paigegail
Acceleration = 1.733 m/s -squared-

That's not what I got for acceleration. I got 2.13 m/s2.

If you get the net acceleration right, then you multiply it by the mass to get the net force, as you described.
 
  • #16
How did you get that for acceleration?
 
  • #17
Originally posted by paigegail
v-squared- = vo-squared- + 2aDelta X
4.163 -squared- = 0 + 2(5)a
17.330569 =10a
a=1.733
You have a typo: the value for v is not 4.16, it's 4.61.
 
  • #18
Thank you soooooooooooooo much for everything! OMG THIS WAS SOOO FRUSTRATING. But thank you sooooo much!


-Paige
 
  • #19
Are we done?
 
  • #20
YES! And I really have to thank you for being patient with my annoyance. I was a little bit annoying and I know it. So thanx.

-Paige
 

1. How do you calculate the horizontal speed of a falling box?

The horizontal speed of a falling box can be calculated by using the formula v = d/t, where v is the velocity, d is the horizontal distance traveled, and t is the time the box takes to fall.

2. What units should be used to express the horizontal speed of the falling box?

The horizontal speed of the falling box should be expressed in meters per second (m/s).

3. Is the horizontal speed of the box affected by air resistance?

Yes, the horizontal speed of the box is affected by air resistance. As the box falls, it will experience air resistance which will slow down its horizontal speed.

4. Can the horizontal speed of the box be greater than its vertical speed?

No, the horizontal speed of the box cannot be greater than its vertical speed. In a free fall, the vertical and horizontal speeds are directly proportional, so if the box has a greater vertical speed, its horizontal speed will also be greater.

5. How can the horizontal speed of the box be used to predict its landing location?

The horizontal speed of the box, along with the vertical speed, can be used to predict the landing location of the box. By knowing the initial horizontal speed and the time it takes to fall, you can calculate the horizontal distance traveled and determine where the box will land.

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