Calculating Work and Kinetic Energy in a Vertical Helicopter Lift

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To calculate the work done by a helicopter lifting a 72.7 kg astronaut 16.9 m with a constant upward acceleration of 7.23 m/s², the correct approach involves determining the total force exerted by the helicopter. This force includes both the gravitational force (mg) and the force required for the upward acceleration (ma). The formula for work is then applied as Work = Force x Distance. The initial attempts at calculating work were incorrect due to miscalculating the force required. The final formula incorporates both components to yield the correct work done by the helicopter.
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A helicopter is used to lift a 72.7 kg astronaut 16.9 m vertically out of the ocean by means of a cable. The astronaut rises with constant upward acceleration of magnitude 7.23 m/s2 until she reaches the helicopter.

a) How much work is done by the helicopter on the astronaut?


W=1/2MV^2=FS


first try: V^2=2as W=mas=8882.9949J wrong
second try: F=mg W=mgh=12052.8603J wrong

someone help me please, thank you so much!
 
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First find the force required to lift the person. Then use work is force times displacement to find the work.
 
is the force mg ?
 
Use Newton's Second Law.

F = m*a

Force of helicopter - mass of astronaut * g = mass of astro * 7.23

Therefore the Force the helicopter must exert is 7.23 m + mg

Work = Force x Distance, so its just (7.23 m + mg) (16.9) Joules. Good luck!
 
yeah, i forgot to add the force of helicopter. thank you man!
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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