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2 kg
Use 10 m/s2 for g

2 Kg Use 10 Ms2 For G class=

Sagot :

Answer:

See below

Explanation:

At point A    the PE = mgh = 2 * 10 * 1 = 20 J

  at point B, all of the PE , 20 J , is converted to Kinetic Energy

KE = 1/2 m v^2

20 = 1/2 (2)(v^2 )

20 = v^2     v = sqrt 20 = 4.47 m/s

for the friction part

 vf = vo t  + 1/2 a t^2      vf = final velocity = 0 (stopped)

                                       vo = original velocity = 4.47 m/s  

                                         a = -1 m/s^2

  0 = 4.47 t + 1/2 (-1) t^2

            - .5t^2  + 4.47 t = 0

                 t ( -.5t+ 4.47) = 0    shows t =  4.47/.5 = 8.9 seconds

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