Asked by April
                A 7.4kg block starts 1.2m above the ground, at the top of a frictionless ramp.  At the bottom of the ramp is a flat stretch of rough (μk = 0.15) ground 3.0m long.  After sliding 3.0m, the ground becomes frictionless, and the block hits a spring (k = 1.6kN/m).
a. Calculate the speed of the block at the bottom of the ramp.
            
        a. Calculate the speed of the block at the bottom of the ramp.
Answers
                    Answered by
            Scott
            
    1/2 m v^2 = m g h
v = √(2 * 9.8 * 1.2)
    
v = √(2 * 9.8 * 1.2)
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