We were given the answer to this puzzler and its 71.96875 N
Yes, that's the same answer I got using the method I described to you.
What I did...
I will show you what I did and then can you please tell me what I'm doing wrong.
¥Äx= -12.96 m
Vo= 0 m/s
a= -9.80 m/s
v©÷ = vo©÷ + 2a¥Äx
v©÷ = 0 + 2a¥Äx
v = vo + at
15.93 = 0 + (-9.80)t
¥Äx = 7.5 + 5= 12.5 m
t = 1.626 s
¥Äx = vt
= 7.688 m/s
What do I do from here and is this really right?
The equations in the last one are a little wack. sorry...
© = squared
¥Äx= delta x
Re: What I did...
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!)
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:
= 10.20 m/s-squared-
Am I going right...because then I get F= ma
F= 10.8 (10.20)
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.
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.
Delta X = 5.0 m
Vx= 4.613 m/s
Delta X = Vx(t)
t= delta x/ vx
= 1.083 s
So do I add 1.083 to the previously establish 1.626 and then get 2.709 for total time?
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.
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.
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?
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)
So is that Fnet? If it is...then what?
How did you calculate the acceleration?
v-squared- = vo-squared- + 2aDelta X
4.163 -squared- = 0 + 2(5)a
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.
How did you get that for acceleration?
You have a typo: the value for v is not 4.16, it's 4.61.
Thank you soooooooooooooo much for everything!!! OMG THIS WAS SOOO FRUSTRATING. But thank you sooooo much!!
Are we done?
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.
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