text: We will assume that air and spring friction is negligible. We will also assume that the mass starts falling with a velocity of zero (it isn't thrown, for example).\n\nFirst, we find the velocity of the mass at the time it strikes the spring. distance = (1/2)gt^2 so:\n4 = (1/2)(9.8)t^2, t = 0.90 sec.\nv = gt = (9.8)(0.90) = 8.85 meters/sec\n\nSo the kinetic energy of the mass when it hits is (1/2)mv^2,\nwhich is (1/2)m(8.85)^2 or 39.2m Joules.\n\nThe work done by gravity as the spring compresses is f*d or\n(9.8)m(1) or 9.8m Joules.\n\nNext, we need to know the work done by the spring as it compresses. This is the integral of force over distance or\nInt(1000LdL). Integrating gives 500 L^2 evaluated from 1 to 0 which is -500 Joules.\n\nFinally, we have our equation. The Kinetic energy of the mass is absorbed by the spring, as is the work done by gravity. So we have :\n\n39.2m + 9.8m = 500\n49m = 500\nm = 10.2 kg.