Hey Physics Experts! I got 2 questions which I did my work, but want to double check with someone, anyone! Q:A rescue helicoper lifts a 79 kg person straight up by means of a cable. The person has an upward acceleration of 0.70m/s^2 and is lifted from rest through a distance of 11m. a)what is the tension in the cable? How much work is done by b)the tension in the cable and c)the person's weight? d) use the work-energy theorem and find the final speed of the person. My work so far: a)Tension in cable sum of Fy=T-W=ma T=ma+W 79kg(0.70m/s^2)+79(9.81m/s^2)= 830.29 N b)Work by tension in cable: W=Fcos(0 degrees)s =830N(1)(11m) =9130J c)work done by person's weight W=Fcos(180)s =774.99N(-1)(11m) =-8525 J d)Final speed KEf=W+KEo 1/2 mv^2=W+1/2mv^2 Vf=square root (2*9133J/79kg) =15.20m/s Any suggesions great. Here's the other one... Q:A particle, starting from point A in the darwing, is projected down the curved runway. upon leaving the runway at point B, the partticle is traveling straight upward reaches a height of 4.00m above the floor before falling back down. ignoring firction and air resistance, find the speed of the particle at point A. I've attached a crude paint drawing. Ok my work: Wnc (non conservative work) = 0 Ef=Eo 1/2mv^2 + mgh= 1/2mVo^2+mgh Since kinetic and potential must be constant then I can cancel some terms so mgh=1/2mVo^2 Vo=square root(2*9.81*(4m-3m)) Vo= 4.43m/s Sounds too simple if you ask me. Somethings in err, comments? Thanks all.