The horizontal surface on which the block (mass 2.0 kg) slides is frictionless. The speed of the block before it touches the spr
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Answer:
6.16 m/s
0.0105 m
Explanation:
Let the ground 0 for potential reference be at where the spring is compress 0.24 m. The the man would jump from a height h = 2.5 + 0.24 = 2.74 m from it. We can apply the law of energy conservation knowing that as the man jumps, his potential energy converts to kinetic energy, then finally to elastic energy:
where m = 80 kg is the man mass, g = 9.81 m/s2 is the gravitational acceleration, h = 2.74 m is the potential distance he travels, k N/m is the spring constant and x = 0.24 is the distance it compresses
Similarly at the position where it compresses by 0.12 m, it's 0.24 - 0.12 = 0.12 m far from ground 0.
When he steps gently, then his gravity force would equal to his spring force