Question

a 1.5 kg object is thrown down with a speed of 10 m/s from the top of a building that is some distance above a platform that is attached to an unstretched spring. the object hits the platform and compresses the spring by 2.6 m before coming to rest. the spring has a stiffness of 256 N/m. how high is the top of the building above the platform?

Answer 1

$\! \! \! \! \! \! \! \! \! $Let h be the height of the building. Taking the platform level as zero reference for gravitational potential energy.$\\\\ $Total energy of the object at the top of the building, $E_{total} = \frac{1}{2}mv^2 + mgh\\\\ $Potential energy of a spring having spring constant k stretched to a distance x from the equilibrium position, $\\\\ PE = \frac{1}{2}kx^2\\\\ $Total energy of the object when 2.6m below the platform,$\\\\ E_{total} = \frac{1}{2}kx^2 - mg(2.6) = \frac{1}{2}\times 256\times 2.6^2 - (1.5\times 9.8\times 2.6)=827.06$J$\\\\ $Applying conservation of energy to the problem, $\\\\ \frac{1}{2}mv^2 + mgh=827.06\\\\ $Height of the building, $\\\\ h = \frac{826.07-\frac{1}{2}mv^2}{mg}= \frac{826.07-\frac{1}{2}\times 1.5\times 10^2}{1.5\times 9.8}=51.09$m$$