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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 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?
one year ago

Arun
18718 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 = $\sqrt$47.5 = 6.9 m/sec

Regards

one year ago
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### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions