- #1

paigegail

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

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- Thread starter paigegail
- Start date

- #1

paigegail

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

- #2

Ambitwistor

- 841

- 1

Yes, that's the same answer I got using the method I described to you.

- #3

paigegail

- 22

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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

paigegail

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The equations in the last one are a little wack. sorry...

© = squared

¥Äx= delta x

© = squared

¥Äx= delta x

Last edited:

- #5

Doc Al

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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:Originally posted by paigegail

Consider horizontal

¥Äx = 7.5 + 5= 12.5 m

distance = 7.5m (do not add the distance the box was pushed!)

- #6

paigegail

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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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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

paigegail

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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

Ambitwistor

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- #10

paigegail

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- #11

Doc Al

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Relax....Originally posted by paigegail

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

paigegail

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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

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How did you calculate the acceleration?Originally posted by paigegail

K...I found acceleration wrong. I realized that.

Acceleration = 1.733 m/s -squared-

- #14

paigegail

- 22

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v-squared- = vo-squared- + 2aDelta X

4.163 -squared- = 0 + 2(5)a

17.330569 =10a

a=1.733

4.163 -squared- = 0 + 2(5)a

17.330569 =10a

a=1.733

- #15

Ambitwistor

- 841

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Originally posted by paigegail

Acceleration = 1.733 m/s -squared-

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

If you get the net acceleration right, then you multiply it by the mass to get the net force, as you described.

- #16

paigegail

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How did you get that for acceleration?

- #17

Doc Al

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You have a typo: the value for v is not 4.16, it's 4.61.Originally posted by paigegail

v-squared- = vo-squared- + 2aDelta X

4.163 -squared- = 0 + 2(5)a

17.330569 =10a

a=1.733

- #18

paigegail

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-Paige

- #19

Doc Al

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Are we done?

- #20

paigegail

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-Paige

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