A child releases himself from a tree branch 50 cm above a platform onto the plat

Question

A child releases himself from a tree branch 50 cm above a platform onto the platform. THe platform is mounted onto a spring. Together, they come to rest 10 cm below the initila position of the platform. a.) Assume the masses ofthe child and the platform are 10 kg and 2 kg, respectively. What is the spring constant ofthe spring? b.) Assume that the platform-spring system is designed such that it would bring a 10 kg child to rest in minimum time without overshooting. What is the damping factor? c.) Find the equation of motion for the platform, choosing upward as the positive direction, the final rest position as the origin of the x-coordinate, and the child's impact instant as t=0. Additionally, treat the child's impact to the platform as a perfect inelastic collision between the child and the platform.

Explanation / Answer

a) h = height above the normal position of the platform = 50 cm = 0.50 m

x = compression of platform = 10 cm = 0.10 m

m = mass of platform = 2 kg

M = mass of child = 10 kg

k = spring constant

using conservation of energy

mgx + Mg(h + x) = (0.5) k x2

2 x 9.8 x 0.10 + 10 x 9.8 (0.50 + 0.10) = (0.5) k (0.10)2

k = 12152 N/m

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