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`        The co-efficient of friction between a rubber tyre and a wet concrete road is 0.5. Find the minimum time in which a car whose initial velocity is 15m/s can come to a stop on such a road.`
2 years ago

MAINAK KHAN
42 Points
```							Coff. of friction= ½initial velocity(u) = 15m/sfinal velocity(v)   = 0 m/sdistance(S) = ?applying work energy theoremWORKall = ∂kwork is done only by friction according to the questionWORKfriction= 1/2mv2 – 1/2mu2-1/2*m*g*S = 0 – ½m(15)2-1/2*g*S = -1/2*(15)2-5*S   = -1/2*225 (since g = 10 m/s2)S = 225/10S = 22.5
```
2 years ago
MAINAK KHAN
42 Points
```							Coefficient of friction = ½initial velocity(u) = 15m/sfinal velocity(v) = 0Acceleration(a) =?Distance(S) = ?FORCE = MASS* ACCELERATION FORCEfriction = ma-1/2*m*g = maa = -5m/s2applying in 3 equation of motionv2 = u2 + 2aS0 = 225 + 2(-5)S-225 = -10*SS= 22.5m
```
2 years ago
Bhavasagar
45 Points
```							Given,u=15m/s^2; coefficient of friction=0.5Net force acting on the body on X axis is f=ma. 0.5mg=ma1/2*10=aa=5m/s^2From,v=u+at. 0=15+(-5)t. t=3s
```
one year ago
Gitanjali Rout
184 Points
```							 Coff. of friction= ½initial velocity(u) = 15m/sfinal velocity(v)   = 0 m/sdistance(S) = ?applying work energy theoremWORKall = ∂kwork is done only by friction according to the questionWORKfriction= 1/2mv2 – 1/2mu2-1/2*m*g*S = 0 – ½m(15)2-1/2*g*S = -1/2*(15)2-5*S   = -1/2*225 (since g = 10 m/s2)S = 225/10S = 22.5
```
one year ago
Gitanjali Rout
184 Points
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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