Recent content by ChrisGeophysics

New on here. Signed up today. I am working towards a B.S. Geophysical Engineering and an M.S. in Geoscience.

Haruspex, you are correct. I made a poor assumption that the graph continued to (0,0) for I had just seen and assigned this question. Thank you. The correct work is 51J as you stated, which results in a final velocity of 14.456 m/s ==> 14.5 m/s.

Actually the answer given is correct when rounding 14.999 to three sig fig. Work is area under curve of force vs position graph. Setting 55 = 1/2mVf^2 - 1/2mVi^2. When working with five sig figs for calculations (this comes from squaring the velocity) your answer will be 14.999 m/s => 15.0 m/s