- #1
Westin
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- 0
Homework Statement
A 27kg box is being pushed across the floor by a constant force [PLAIN]https://s3.lite.msu.edu/adm/jsMath/fonts/cmsy10/alpha/144/char68.png96,0,0[PLAIN]https://s3.lite.msu.edu/adm/jsMath/fonts/cmsy10/alpha/144/char69.pngN. The coefficient of kinetic friction for the table and box is 0.14. At t=6s the box is at location [PLAIN]https://s3.lite.msu.edu/adm/jsMath/fonts/cmsy10/alpha/144/char68.png7,6,7[PLAIN]https://s3.lite.msu.edu/adm/jsMath/fonts/cmsy10/alpha/144/char69.pngm, traveling with velocity [PLAIN]https://s3.lite.msu.edu/adm/jsMath/fonts/cmsy10/alpha/144/char68.png9.5,0,0[PLAIN]https://s3.lite.msu.edu/adm/jsMath/fonts/cmsy10/alpha/144/char69.pngm[PLAIN]https://s3.lite.msu.edu/adm/jsMath/fonts/cmmi10/alpha/144/char3D.pngs.
What is its velocity at 8.6s?
v[PLAIN]https://s3.lite.msu.edu/adm/jsMath/fonts/cmmi10/alpha/144/char7E.png=[PLAIN]https://s3.lite.msu.edu/adm/jsMath/fonts/cmex10/alpha/144/char1C.png 15.1736 m/s , 0 m/s , 0 m/s https://s3.lite.msu.edu/adm/jsMath/fonts/cmex10/alpha/144/char1D.png
What is its position at 8.6s? (Hint: Because the time step is so large, make sure to use the average velocity for this prediction.)
d[PLAIN]https://s3.lite.msu.edu/adm/jsMath/fonts/cmmi10/alpha/144/char7E.png=[PLAIN]https://s3.lite.msu.edu/adm/jsMath/fonts/cmex10/alpha/144/char1C.png , , https://s3.lite.msu.edu/adm/jsMath/fonts/cmex10/alpha/144/char1D.png
Homework Equations
[/B]
Kinematic Equations
Position Update Equations
Friction Equations
The Attempt at a Solution
How I found velocity..
Force of friction = .14*27*9.8 = 37.044
acceleration = (96N - 37.044)/27 = 2.18m/s^2
Final Velocity = Vi + 2.18(2.6) = 15.1736m/s
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