Find the potential and kinetic energy of the box

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Homework Help Overview

The problem involves calculating the potential and kinetic energy of an 8.63 kg box sliding down a frictionless incline at a specific time. The incline has an angle of 20 degrees, and the box starts from rest at a height of 24.6 m.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to calculate the kinetic energy using the formula .5mv^2 after determining the velocity based on acceleration. They also calculate potential energy using mgh and then subtract kinetic energy from potential energy. Some participants question the correctness of the calculations and the input method into the online system.

Discussion Status

Participants are discussing the potential reasons for the discrepancy in the original poster's results when entering them into the UTexas system. There is no explicit consensus on the source of the error, but suggestions have been made regarding significant figures and the possibility of a system error.

Contextual Notes

The original poster mentions that the online program allows for a 1% error, and they express uncertainty about the input requirements. There is also a mention of being snowed in, indicating a potential delay in seeking further assistance from the teacher.

Houyhnhnm
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A 8.63 kg box slides down a long, frictionless incline of angle 20 degrees. It starts from rest at time t = 0 at the top of the incline a height 24.6 m above ground.
Find the potential and kinetic energy of the box at t = 1.8s. Answer in units of J.

acceleration is gsin20
so I got velocity
by gsin20 x 1.8 seconds
and then I used .5mv^2 to find KE
and then for PE
I found PE at the highest point
which is mgh
and then subtracted KE from that

I got 1925.2569 J for PE and 157.3863974J for KE (ignore sig figs)
I entered this into UTexas and got it wrong.
Please help.
 
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Maybe I've had too much to drink this eveing but I can't see where you went wrong.
 
Yeah neither can I
 
Where is this UTexas thing?

Doug
 
Originally posted by Houyhnhnm
I got 1925.2569 J for PE and 157.3863974J for KE (ignore sig figs)
I entered this into UTexas and got it wrong.
Please help.
Your method is fine. I assume you are entering your answer into some online system? If so, they can be picky; it might not tell you you are correct unless you give exactly the answer it expects. Try varying the number of significant figures you enter. (I hope you didn't enter all 8-10 "significant" figures!)
 
This UTexas thing is at hw.utexas.edu. It's an online program that my physics teacher uses to torture us. Doc Al, the program specifically says "Enter what your calculator says" and it allows a 1% error anyways so I'm pretty sure that's not the problem.
 
It's always possible that the program might contain an error. Have you asked your teacher about this problem?
 
Heh...we've been snowed in for 2 days...problem has some time to fester...
 

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