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- #1

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- #1

- Mar 1, 2012

- 935

show your work on this problem, please

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- #3

I don't understand how to approach this problem.show your work on this problem, please

m=50kg, it says slowing down from 10m/s, so it's decelerating, u=10m/s

Using F=m×a, 25-(0.3×500)=50a, a=-2.5m/s^2. By taking the total distance travelled under friction, I get v=5.24m/s, so making the assumption that the box comes to halt instantly, I got s= 1.05m when subtracted from total distance I get 13.5 m but the ans is 13m

- Mar 1, 2012

- 935

$v_f^2 = 10^2 - 2(3.5) \cdot \Delta x_1 \implies v_f^2 = 100 - 7\Delta x_1$

after the string breaks ...

$0^2 = (100 - 7\Delta x_1) - 2(3) \cdot \Delta x_2$

$\Delta x_2 = 14.5 - \Delta x_1 \implies 100 - 7\Delta x_1 = 6(14.5 - \Delta x_1) \implies \Delta x_1 = 13 \, m$

poorly worded problem. imho.

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- #5

Thank you so so much!!!!

$v_f^2 = 10^2 - 2(3.5) \cdot \Delta x_1 \implies v_f^2 = 100 - 7\Delta x_1$

after the string breaks ...

$0^2 = (100 - 7\Delta x_1) - 2(3) \cdot \Delta x_2$

$\Delta x_2 = 14.5 - \Delta x_1 \implies 100 - 7\Delta x_1 = 6(14.5 - \Delta x_1) \implies \Delta x_1 = 13 \, m$

poorly worded problem. imho.