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`        A block of mass 15 kgis placed on inclined plane when an angel of inclination is 30 the block just begin to slide the plane what is the sliding fricition`
one year ago

Arun
24740 Points
```
first of all 1 kg.wt means a force that is required to accelerate a mass of 15 kg with an acceleration of approx 9.8 m/s2.
here the block is just going to slide down the wedge..so friction force needed to stop the block is mg*sin30.
since, mass  = 15kg, force required  = 15*g*sin30 newton
= 7.5*g  newton
= 7.5 kg.wt …............... {since 1 kg.wt = 1*g newton, where g = 9.8 m/s2 }

```
one year ago
Khimraj
3007 Points
```							In equilibrium,F=mg sin 30°…………(1)R=mg cos 30°…………(2)On dividing ( 1) by (2), we getF/R= mg sin 30°/mg cos 30°or, friction coefficient= tan 30°.
```
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