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Mechanics

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23 Sep 2007 20:45:48 IST

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sir plzz ans my ques.

None

a car starts from rests on a half kilometer long bridge.the cofficient of friction between the tyre and the road is 1.0 show that one cannot drive through it in less than 10s.

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ramyani chakrabarty

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Joined: 22 Apr 2007

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23 Sep 2007 21:04:37 IST

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Let us consider the limiting case.
The car moves forward becoz of the frictional force.
So
umg = ma   ( u = coef of friction, m = mass, a = accln)

                     a = g = 10 m/s/s

                    using the equn, S = xt + (1/2) at²   ( x= initial vel = 0)

                                   Here S = 500 m

                                    we get t= 10 s

This is the case when friction was considered in the dirn of motion.

Mahesh Jain

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23 Sep 2007 21:04:54 IST

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first of all i m not sir but herez ur solution

let force acting on car for acceleration be F and Frictional Force is umg

u=coefficient of friction

now acceleration of car provided by the F=MA

M=Mass of the Car

Now retardation provided by the Friction is ug=a

Relative least acceleration of the car is A-a=ug

Now applying the s=vt+1/2 at^2

v=0 as it starts from rest

convert distance in metres

and put in the equation

1/2 *1000=1/2 ug t^2

1000=10 t^2

100=t^2

10=t

t=10

rate if u r satisfied

shami pratap

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Joined: 5 Aug 2007

Posts: 38

23 Sep 2007 21:15:43 IST

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car gets its accereration from reaction of earth( deck of bridge). Friction is helpful for us! The accelaration a (if coeff.of frict.=mu), then a will be given by mu.mg/m=mu.g=1xg=g. Now use 2nd law.
500=0+0.5gt.t => t=sqrt(100)=10 sec.

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