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A box of mass 8.0 kg slides at a speed of 10 ms−1 across a smooth level floor before itencounters a rough patch of length 3.0 m. The frictional force on the box due to thispart 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?

ROHIT SHARMA , 6 Years ago
Grade 12th pass
anser 1 Answers
Arun

Last Activity: 6 Years ago

 
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)

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