A 110kg box, starting from rest, is pulled across a floor with a constant horizontal force of 350N. For the first 15m the floor is frictionless, and for the next 15m the coefficient of friction is 0.30. What's the final speed of the box?
2006-10-11 10:38:00 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Physics
The box accelerates at 350/110 m/s^2 for 15 m. Then it accelerates at 350-9.81*110*0.3/110 m/s^2 for another 15 m. You can do the math
2006-10-11 10:43:37 · answer #1 · answered by kirchwey 7 · 0⤊ 0⤋
first find out acceleration (A) for the first 15m, using formula
X=Xi+VT+0.5AT^2
then find T, plug it in to V=Vi+AT, you get Vf for the first 15.
Do the same thing with second 15m, but Vi=9.73 as the result for Vf from the first part is 9.73
Do the same thing, you get
12.425m/s?
I suck at physics, so i could be wrong
2006-10-11 17:45:17 · answer #2 · answered by Yue J 3 · 0⤊ 0⤋