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Grade: 12th pass
        A box of mass 8.0 kg slides at a speed of 10 ms−1 across a smooth level floor before it encounters a rough patch of length 3.0 m. The frictional force on the box due to this part of the floor is 70 N. What is the speed of the box when it leaves this rough surface? What length of the rough surface would bring the box completely to rest?
5 months ago

Answers : (1)

Arun
11696 Points
							
Dear student
 
a = F/m
a = 70/8 = 8.75 m/sec²
v² = u² + 2as
v² = 10² + 2 (-8.75)*3
v² = 100 - 52.5 = 47.5
v = \sqrt47.5 = 6.9 m/sec
 
Regards
Arun (askIITians forum expert)
5 months ago
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