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Physics Mechanics- friction

Shah 72

Member
Apr 14, 2021
193
A box of mass 50 kg is slowing down from 10 m/s on a rough horizontal ground. The coefficient of friction between the box and the ground is 0.3. To start with, the box is being slowed by a string providing a tension of 25N horizontally. Then the string breaks and the box comes to a halt under friction alone after a total distance of 14.5m. Find how far the box travelled before the string broke.
 

skeeter

Well-known member
MHB Math Helper
Mar 1, 2012
935
show your work on this problem, please
 

Shah 72

Member
Apr 14, 2021
193
show your work on this problem, please
I don't understand how to approach this problem.
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
 

skeeter

Well-known member
MHB Math Helper
Mar 1, 2012
935
To get 13m as the solution, the 25N applied force acts in the same direction as the friction force before the string breaks.

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

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.
 

Shah 72

Member
Apr 14, 2021
193
To get 13m as the solution, the 25N applied force acts in the same direction as the friction force before the string breaks.

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

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.
Thank you so so much!!!!