Question

5. The picture shows a mass that is attached to a spring with constant, K = 500 N/m(20 points)

Answer 1

Answer:

$\text{(a)}\\\text{Given that current energy at starting point }h_{1} \text{ is}:\\E_{h1}=mgh+\frac{1}{2}kx^2\\\text{Plug in height and mass: }E_{h1}=10g+250(0.8)^2=10g+160=258J\\\text{Using Conservation of Energy: }E_{h1}=E_{h2} \\\therefore 258J=mg(h_{2}-h_{1})+\frac{1}{2}mv^2\\=258J=10g+v^2\\=160=v^2\\\therefore v=40m/s \text{ at }h_{2}$$\text{(b)}\\\text{Again, apply conservation of energy: } 258J=mg(h_{3}-h_{2})+\frac{1}{2}mv^2\\\text{Plugging in: }258J=16g+v^2 \\\text{ Simplify: }101.2=v^2\\\therefore v=10.05m/s \text{ at } h_{3}$