I don't understand how to approach this problem.skeeter said:show your work on this problem, please
Thank you so so much!skeeter said: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.
The concept of friction refers to the resistance force that occurs when two surfaces come into contact with each other. In this case, the friction between the box and the surface it is sliding on causes it to slow down and eventually come to a halt.
The mass of the box affects its motion and the distance it travels because it determines the amount of force needed to accelerate or decelerate the box. In this scenario, the 50kg mass of the box requires a greater amount of force to overcome the friction and bring it to a halt, resulting in a longer distance traveled.
Yes, the distance traveled by the box can be changed by altering its mass. A heavier box would require more force to overcome the friction and come to a halt, resulting in a longer distance traveled. On the other hand, a lighter box would require less force and therefore travel a shorter distance before coming to a halt.
Yes, there are ways to reduce the friction between the box and the surface it is sliding on. One way is to use a lubricant such as oil or grease to create a slippery layer between the surfaces. Another way is to use smoother surfaces or materials that have a lower coefficient of friction.
The distance traveled by the box is directly proportional to its initial speed. This means that the faster the box is moving, the longer the distance it will travel before coming to a halt due to friction. However, other factors such as mass and surface friction can also affect the distance traveled.