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

## Answers and Replies

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

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©÷ = 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...
¥Äx= delta x

Last edited:
Mentor

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 cuz that would make this much easier.

Mentor
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?

Ambitwistor
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.

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.

Mentor
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?

Mentor
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

Ambitwistor
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.

paigegail
How did you get that for acceleration?

Mentor
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

Mentor
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