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

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The discussion centers on the physics of a 12-kg box sliding across a rough surface with a frictional force of 72N. The work done by the frictional force is calculated using the formula W = F × d, resulting in 216 Joules for the 3 m rough section. The box's velocity upon exiting the rough surface is determined to be 6 m/s, and the length of rough surface required to bring the box to a complete stop is calculated to be 7.5 m.

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odie
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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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