I have a test tomorrow, and I got a list of practice problems, but for the life of me I can't figure out how to solve this problem:quot;A crane lifts a 50-kg crate so that the crate's speed increases from 0 m/s to 5 m/s over a vertical distance of 10.0 mDraw a bar chart representing this process of the crane-crate systemWhat is the force that the crane exerts on the crate? Use the problem solving strategy and your knowledge of conservationOmit unitsRound to the tenth.quot; if possible could someone walk me through the final of the bar chart? I know there is work done, but would the final Grav pe be more or less than the final KE?
work-energy theorem: Δ k.e net work done net force x distance moved in force direction(line of action) the same can also be expressed as Δ k.e + Δ p.e F x 10 - (i) [conservation of Energy] F is the force applied via the crane on the crate Δ k.e final k.e - initial k.e ?m5? - ?m0? ?50 x 25 625 J Δ p.e final p.e - initial p.e mg(10) - mg0 50 x 9.81 x 10 4905 J from (i) 625 + 4905 10F F 5530/10 553 N also final p.e final k.e 4905 625 perhaps you can draw bars representing the proportional increase of p.e and k.e at various instants like for example when the velocity of the crate is 3 m/s, let the vertical distance be x metres from (i) 504.5 + mgx 553x 225 x[553 - 50(9.81)] [553 - 490.5]x 225 or x 225 / 62.5 3.6 m therefore Δ p.e 50 x 9.81 x 3.6 1765.8 J vs Δ k.e 225 J hope this helps
