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Puzzler Answer

  • Thread starter paigegail
  • Start date
  • #1
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We were given the answer to this puzzler and its 71.96875 N
 

Answers and Replies

  • #2
841
1
Yes, that's the same answer I got using the method I described to you.
 
  • #3
22
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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
22
0
The equations in the last one are a little wack. sorry...
© = squared
¥Äx= delta x
 
Last edited:
  • #5
Doc Al
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1,169


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
22
0
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 cuz that would make this much easier.
 
  • #7
Doc Al
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44,910
1,169
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
22
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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
841
1
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
22
0
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
Doc Al
Mentor
44,910
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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
22
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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
Doc Al
Mentor
44,910
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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
22
0
v-squared- = vo-squared- + 2aDelta X
4.163 -squared- = 0 + 2(5)a
17.330569 =10a
a=1.733
 
  • #15
841
1
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
22
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How did you get that for acceleration?
 
  • #17
Doc Al
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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
22
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Thank you soooooooooooooo much for everything!!! OMG THIS WAS SOOO FRUSTRATING. But thank you sooooo much!!


-Paige
 
  • #19
Doc Al
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Are we done?
 
  • #20
22
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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
 

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