During the fall she loses 6 meters of altitude.
That means she looses m g h = 30 * 10 * 6 = 1800 Joules of potential energy
That energy went into stretching the spring
so
1800 Joules of work done
energy in spring = (1/2) k x^2 where x is spring compression or tension
so
(1/2) k (1)^2 = 1800 solve for k
A little girl of mass 30kg is jumping straight up and down on a stretched trampoline. During a single jump, she reaches a maximum height of 5.0m above the horizontal surface of the trampoline and deeps to 1.0m below the surface. Use the acceleration due to g=10m/s^2 and treat the girl as a point particle.
(a) Calculate the work done on the girl by the elastic force of the trampoline during a single fall.
(b) Now treat the trampoline as a simple spring that obeys the Hooke's law and obtain it's spring constant.
1 answer
1 answer