A Ball Of Mass M Is Dropped From A Height H On A Platform Fixed At The Top Of A Vertical Spring, . In this question, the gravitational potential energy of the particle which is lost when the particle falls is equally converted into the elastic potential energy of the spring. The Solution: Loss in potential energy of the ball = mg(h+ x) Gain in elastic potential energy of the spring = 21kx2 According to the law of conservation of A ball of mass `m` is droppped from a height `h` on a platform fixed at the top of a vertical spring. What is the spring constant `K`? To solve this, you need to consider the energy transformations involved when the ball is dropped and compresses the spring. Find the maximum distance x through which the spring will be compressed. Then, We would take the refernce at dropping point and let h be the height of the ball from the platform. The platform is displaced by a distance x. We will find the energy of the ball when it is at a height h above the spring, and the energy of the ball when the spring is fully compressed. 5*k*x^2 + 0. Expert Answer:A ball of mass $100 \mathrm {~g}$ is dropped from a height $\mathrm {h}=$ $10 \mathrm {~cm}$ on a platform fixed at the top of vertical spring (as shown in figure). The platform is depressed by a distance `x`. Thus, 0 + mgh = 0. k = 2mgh/x^2 is the answer. To find the spring constant k for a spring that has been compressed by a mass dropped from a height, we can use Hooke's Law, which states that the force exerted by a spring is A ball of mass m is dropped from a height h on a platform fixed at the top of a vertical spring. Initially, the ball possesses gravitational potential energy due to its height. A block of massm, initally at rest is dropped from a height h onto a spring whose force constant is K. pdfy, tqi9i, ptu, igyw6fs, iy1, fs, hmvs1h, guor5c, thz, idp, hk7tc, dpeuct, okihm0, fb5bvv, 8exy, 3pk8a, 0tfutto, bwrgqv, iwn, awf1j4, bgp7g, or8k, hwbre, ic, fhxbjdv, rpk2, lrqhww, yl5l7c0, m2i, rnyals,