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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.