You push a 45 kg wooden box across a wooden floor at a constant speed of 1.0 m/s. The coefficient of kinetic friction is 0.25. Now you doubl

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1. Answer:

          t = 0.408 s

   Explanation:

   In this exercise we will use Newton’s second law to find the acceleration and the kinematic relationships to find the time.

   We use Newton’s second law, for the first condition where the acceleration is zero

   X axis

              F – fr = 0

   Y Axis  

             N- W = 0

   The expression for friction force is

           fr = μ N

          fr = μ mg

   We substitute

         F = μ m g

   This is the value for the applied force, now we apply the second condition, in this case the system has an acceleration

   X axis

             2F- fr = ma

            a = (2F -fr) / m

   Axis y

            N -W = 0

   The friction force has the expression

           fr = μ N

           fr = μ W

   We substitute

          a = (2F – μ m g) / m

          a = (2 μ mg – μ mg) / m

          a = μ g (2-1) = μ g

   We calculate

         a = 0.25  9.8   1

         a = 2.45 m / s²

   With this acceleration value we use the kinematics ratio to find the time

          v = v₀ + a t

   The initial speed is 1 m / s and the final speed is 2 m / s

          t = (v-v₀) / a

          t = (2-1) / 2.45

          t = 0.408 s

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