- #1
DavidAp
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There's a cord attached to a cart that can slide along a frictionless horizontal rail aligned along an x axis. The left end of the cord is pulled over a pulley, of negligible mass and friction and at cord height h = 1.3 m, so the cart slides from x1 = 5.0 m to x2 = 1.0 m. During the move, the tension in the cord is a constant 29.0 N. What is the change in the kinetic energy of the cart during the move?
Answer 102 J
I have no idea how to approach this problem. I know the equation W = ΔK + ΔU and since we are looking for ΔK I was thinking that I can rearrange the equation to look like,
W - Uf + Ui = ΔK.
However, this raises the question: what's the Work done in this equation? What is the potential energy? From these question I thought I would solve for Work using W = ∫Fdr so,
W = 29N(4m) = 116 Nm
However, something about the way I solved for Work doesn't feel right. I'm still relatively new to Work and am having difficulty putting two and two together when it comes to this topic. If someone can guide me on this problem and, more importantly, explain to me why you chose the equations you did I would greatly appreciate it!
Thank you for taking the time to review my question.
Answer 102 J
I have no idea how to approach this problem. I know the equation W = ΔK + ΔU and since we are looking for ΔK I was thinking that I can rearrange the equation to look like,
W - Uf + Ui = ΔK.
However, this raises the question: what's the Work done in this equation? What is the potential energy? From these question I thought I would solve for Work using W = ∫Fdr so,
W = 29N(4m) = 116 Nm
However, something about the way I solved for Work doesn't feel right. I'm still relatively new to Work and am having difficulty putting two and two together when it comes to this topic. If someone can guide me on this problem and, more importantly, explain to me why you chose the equations you did I would greatly appreciate it!
Thank you for taking the time to review my question.