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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.
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.
