1)a body of mass 5 kg is under the action of 50N on the horizontal surface .if the coefficient of friction in bet the surface is 1 the distance it travels in 3 seconds is 2)a block of weight 100N is pushed by a force F on a horizontal rough plane moves with an acceleration 1m/s2^ the coefficient of friction is 3) a horizontal force applied on a body on a rough horizontal surface produces an acceleration A,if coefficient of friction bet body and surface which is m is reduced to m/3,the acceleration increases by 2 units.the value of m is PLZZZZZ PUT THE QUE NO. BEFORE ANSWERING

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2014-10-05T10:05:53+05:30

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1) 

    Frictional force on the body = μ m g = 1 * 5 kg * g = 5 g   N
    Net force acting on the body = 50 N - friction force = 50 N - 5 g 

If g = 10 m/s², then net force = 50 - 50 N = 0. So acceleration is 0. Velocity = 0.
The body stays at rest.

If g = 9.81 m/s² then net force = 50 - 5 * 9.8 = 1 N
acceleration = net force / mass = 1 / 5 = 0.2 m/sec²

Distance traveled in 3 seconds is S = u t + 1/2 a t² = 0 + 1/2 * 0.2 * 3² = 0.9 meters

2) 

Net force = mass * acceleration = 100N/ g * 1 m/s² = F - μ (m g) = F - μ 100 N

    μ =  [ F - 100 / g ] / 100
 
        If g = 10 m/s²,        μ = (F - 10)/100
           If g = 9.81 m/s²,     μ = ( F - 10.19)/100

3)

           m A = net force = Force applied F - friction μ m g
               A = (F - μ m g)/m = F/m - μ g   --- equation 1
     
acceleration increases by 2 units, if μ is reduced to 1/3μ.
              A + 2 = F/m - (μ/3) g        ---  equation 2

         ( F/m - μ g)  + 2 =  F/m - μ g /3
             2 μ g /3 = 2 

                μ = 3/g     = 0.3  if g =10m/s²   or   0.306,   if g = 9.81 m/s²

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