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`        a force F=5t is applied on a block of mass 10 kg at rest on a rough horizontal surface [co effiecient of friction=.2]what is its speed at t=20 seconds?`
2 years ago

Arun
23328 Points
```							Dear Student  acceleartion = (F-uN)/m =(5t – 0.2 * 100)/10 = (5t – 20)/2 m/s^2a = dv/dt =(5t – 20)/2 m/s^2dv = [(5t – 20)/2]dtintegerating both sides,v = 5t^2/4 – 10tat t=20s V = 500 - 200 = 300 m/sec RegardsArun (askIITians forum expert)
```
2 years 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