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Grade: 9
        
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

Answers : (5)

MAINAK KHAN
42 Points
							
Coff. of friction= ½
initial velocity(u) = 15m/s
final velocity(v)   = 0 m/s
distance(S) = ?
applying work energy theorem
WORKall = ∂k
work is done only by friction according to the question
WORKfriction= 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 = 225/10
S = 22.5 
 
2 years ago
MAINAK KHAN
42 Points
							
Coefficient of friction = ½
initial velocity(u) = 15m/s
final velocity(v) = 0
Acceleration(a) =?
Distance(S) = ?
FORCE = MASS* ACCELERATION 
FORCEfriction = ma
-1/2*m*g = ma
a = -5m/s2
applying in 3 equation of motion
v2 = u2 + 2aS
0 = 225 + 2(-5)S
-225 = -10*S
S= 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= ½
initial velocity(u) = 15m/s
final velocity(v)   = 0 m/s
distance(S) = ?
applying work energy theorem
WORKall = ∂k
work is done only by friction according to the question
WORKfriction= 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 = 225/10
S = 22.5 
one year ago
Gitanjali Rout
184 Points
							
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one year ago
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