Asked by Harold
                A 50 gram block is placed on a frictionless ramp 2 meters above the ground. It slides down the ramp  and compresses a 140 N/m spring at the bottom. How far will the spring be compressed when the block is stopped by the spring?
say 1.7 m
            
        say 1.7 m
Answers
                    Answered by
            drwls
            
    Use conservation of energy. The loss of gravitational potential energy will equal the potential energy stored in the spring. 
0.05 kg * 9.8 m/s^2 * 2 meters = (1/2) k X^2
X^2 = 0.014m^2
X = 0.12 m
I have neglected the gravitational potential energy change during the small spring compression . To include that effect, I would have to know the ramp angle
    
0.05 kg * 9.8 m/s^2 * 2 meters = (1/2) k X^2
X^2 = 0.014m^2
X = 0.12 m
I have neglected the gravitational potential energy change during the small spring compression . To include that effect, I would have to know the ramp angle
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