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PHYSICS help!
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 workenergy theorem and find the final speed of the person.
My work so far:
a)Tension in cable
sum of Fy=TW=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*(4m3m))
Vo= 4.43m/s
Sounds too simple if you ask me.
:confused: :confused: