Figure 7-42 shows a cord attached to a cart that can slide alonga frictionless h

Question

Figure 7-42 shows a cord attached to a cart that can slide alonga frictionless horizontal rail aligned along an x axis.The left end of the cord is pulled over a pulley, of negligiblemass and friction and at cord height h = 1.20 m, so thecart slides from x1 = 3.0 m tox2 = 1.0 m. During the move, the tension in thecord is a constant 25.0 N. What is the change in the kinetic energyof the cart during the move?

I know the answer is 41.7 J, but I dont know how toget it. Any help is appreciated.

Figure 7-42 shows a cord attached to a cart that can slide alonga frictionless horizontal rail aligned along an x axis.The left end of the cord is pulled over a pulley, of negligiblemass and friction and at cord height h = 1.20 m, so thecart slides from x1 = 3.0 m tox2 = 1.0 m. During the move, the tension in thecord is a constant 25.0 N. What is the change in the kinetic energyof the cart during the move? I know the answer is 41.7 J, but I dont know how toget it. Any help is appreciated.

Explanation / Answer

change in kinetic energy = work done on block along thesurface = Tension * displacement * cosine of anglebetween rope and the surface.    If x be distance from the pulley along thesurface, displacement = x1-x small displacement = -dx So, dE = T*(-dx)*cos Now, tan = h/(x) So, cos = (x)/[h2 +x2] So, dE = -T*x/[h2 + x2]dx Now, to get E = change in energy, integrate dE from x=3 to x=1. So, you get E = 41.7 J (approx) You may like to rather represent x in the form ofh/tan = hcot    then,    dx = -hcosec2d    dE = T*cot*cosec2d =(T/sin3)*cosd    => E = -T/2sin2 from = tan-1(h/3) to tan-1(h/1)   

Related Questions

Navigate

• Browse (All)
• Browse F
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.