A helicopter lifts a 45 kg astronaut 29 m vertically from the ocean by means of

Question

A helicopter lifts a 45 kg astronaut 29 m vertically from the ocean by means of a cable. The acceleration of the astronaut it g/10. How much work is done on the astronaut by the force from the helicopter? How much work is done on the astronaut by her weight? What it the kinetic energy? What it the speed of the astronaut just before she reaches the helicopter? A proton (mass m = 1.67 times 10^-27 kg) is being accelerated along a straight line at 2.60 times 10^15 m/s^2 in a machine. The proton has an initial speed of 2.70 times 10^2 m/s and travels 4.3 cm what is its speed? (Give your answer to at least four significant figures.) What is the increase in its kinetic energy?

Explanation / Answer

Here

m = 45 Kg

h = 29 m

a) work done by the force on helicopter = h * (g + a) * m

work done by the force on helicopter = 29 * 45 * (9.8 + 9.8/10)

work done by the force on helicopter = 14067.9 J

b) for the work done by weight

work done by weight = - m * g * h

work done by weight = -29 * 45 * (9.8 )

work done by weight = -12789 J

c) kinetic energy = 14067.9 - 12789

kinetic energy = 1279 J

d) let the speed is v

0.5 * 45 * v^2 = 1279

solving for v

v = 7.54 m/s

the speed is 7.54 m/s

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