What Happens to the Box's Velocity on a Rough Surface?

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The box, with a mass of 12 kg and an initial speed of 10 m/s, encounters a rough surface where it experiences a frictional force of 72 N over a distance of 3 m. The work done by the frictional force can be calculated using the formula W = F * d, resulting in 216 J. To determine the box's velocity after leaving the rough surface, the work-energy principle can be applied, factoring in the initial kinetic energy and the work done against friction. For the box to come to a complete stop, additional calculations are needed to find the distance required to overcome the initial kinetic energy with the same frictional force. The discussion emphasizes the application of physics principles to solve for work, velocity, and stopping distance on a rough surface.
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a box of mass 12-kg slides at a speed of 10m/s across a smooth level floor, where it enters a rough portion 3 m in length. in the rough portion, the box experiences a horizontal frictional force of 72N.

a) How much work is done by the frictional force?
b) what is the velocity of the box when it leaves the rough surface?
c) what length of rough surface brings the box completely to rest?

thats all, don't ask for more cause there is nothing more to add
 
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What did you try?
 
well i found out that we have to use W=(F)(d) (still not sure)
but i have no ideas in how to solve b) and c)
 
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