Calculate the horizontal speed of the box as it falls

[paigegail]

We were given the answer to this puzzler and its 71.96875 N

 

[paigegail]

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?

 

[paigegail]

The equations in the last one are a little wack. sorry...
© = squared
¥Äx= delta x

 

[Doc Al]

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!)

 

[paigegail]

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.

 

[Doc Al]

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.

 

[paigegail]

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?

 

[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.

 

[Doc Al]

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?

 

[paigegail]

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?

 

[Doc Al]

Originally posted by paigegail
K...I found acceleration wrong. I realized that.
Acceleration = 1.733 m/s -squared-

How did you calculate the acceleration?

 

[paigegail]

v-squared- = vo-squared- + 2aDelta X
4.163 -squared- = 0 + 2(5)a
17.330569 =10a
a=1.733

 

[paigegail]

How did you get that for acceleration?

 

[Doc Al]

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.

 

[paigegail]

Thank you soooooooooooooo much for everything! OMG THIS WAS SOOO FRUSTRATING. But thank you sooooo much!


-Paige

 

[Doc Al]

Are we done?

 

[paigegail]

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