# PhysicsMechanics- friction

#### Shah 72

##### Member
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.

MHB Math Helper

#### Shah 72

##### Member
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
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
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!!!!