Asked by Dontrielle
                A small block with mass 0.200 kg is released from rest at the top of a frictionless incline. The block travels a distance 0.440 m down the incline in 2.00 s. The 0.200 kg block is replaced by a 0.400 kg block. If the 0.400 kg block is released from rest at the top of the incline, how far down the incline does it travel in 2.00 s? 
 
            
            
        Answers
                    Answered by
            Henry
            
    0.200 kg block:
d = 0.5a*t^2 = 0.44 m.
0.5a*2^2 = 0.44
2a = 0.44
a = 0.22 m/s^2.
0.400 kg block:
d = 0.5a*t^2 = 0.5 * 0.22*2^2 = 0.44 m
    
d = 0.5a*t^2 = 0.44 m.
0.5a*2^2 = 0.44
2a = 0.44
a = 0.22 m/s^2.
0.400 kg block:
d = 0.5a*t^2 = 0.5 * 0.22*2^2 = 0.44 m
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