Work done on an object being lifted

In summary, the conversation discusses how to calculate the work done on a 72 kg astronaut being lifted 10 m vertically from the ocean by a helicopter and the astronaut's kinetic energy and speed just before reaching the helicopter. The acceleration of the astronaut is g/16 and the question involves finding the work done by the force from the helicopter and the gravitational force, as well as the astronaut's kinetic energy and speed. One participant mistakenly assumes that the upward force from the helicopter is mass times g/16, and another participant points out that the acceleration of the astronaut is not solely due to the helicopter's force. The conversation ends with the participant realizing their mistake and the expert summarizer noting the main points of the discussion.
  • #1
Dan Feerst
12
0

Homework Statement



A helicopter lifts a 72 kg astronaut 10 m vertically from the ocean by means of a cable. The acceleration of the astronaut is g/16. How much work is done on the astronaut by (a) the force from the helicopter and (b) the gravitational force on her? Just before she reaches the helicopter, what are her (c) kinetic energy and (d) speed?


Homework Equations



I have only attempted parts a and b so far, but I can't get a correct answer. I would assume, with a for example that
W=72*10*<9.8/16)*cos(0)=441
I have been trying the same tactic with b. what am I doing wrong?



The Attempt at a Solution

 
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  • #2
Dan Feerst said:
I have only attempted parts a and b so far, but I can't get a correct answer. I would assume, with a for example that
W=72*10*<9.8/16)*cos(0)=441
I have been trying the same tactic with b. what am I doing wrong?

Hi Dan! :smile:

The upward force from the helicopter isn't mass times g/16 … if it was, the astronaut would go down. :wink:
 
  • #3
Wait. Are you saying the question is wrong?






Dan Feerst said:

Homework Statement



A helicopter lifts a 72 kg astronaut 10 m vertically from the ocean by means of a cable. The acceleration of the astronaut is g/16. How much work is done on the astronaut by (a) the force from the helicopter and (b) the gravitational force on her? Just before she reaches the helicopter, what are her (c) kinetic energy and (d) speed?


Homework Equations



I have only attempted parts a and b so far, but I can't get a correct answer. I would assume, with a for example that
W=72*10*<9.8/16)*cos(0)=441
I have been trying the same tactic with b. what am I doing wrong?



The Attempt at a Solution

 
  • #4
No …
The acceleration of the astronaut is g/16

… what makes you think that all comes from the helicopter?
 
  • #5
Not carefully reading the question apparently. :)
 

What is work done on an object being lifted?

Work done on an object being lifted is the energy required to move an object against the force of gravity. It is the product of the force applied to lift the object and the distance the object is lifted.

How is work done on an object being lifted calculated?

The work done on an object being lifted is calculated by multiplying the force applied to lift the object by the distance the object is lifted. This can be represented by the equation W = Fd, where W is work, F is force, and d is distance.

Does lifting an object always require work to be done?

Yes, lifting an object always requires work to be done because it involves moving the object against the force of gravity. Even if the object is lifted at a constant speed, work is still being done to counteract the force of gravity.

What factors affect the amount of work done on an object being lifted?

The amount of work done on an object being lifted is affected by the force applied to lift the object, the distance the object is lifted, and the weight of the object. The greater the force applied or the distance lifted, the more work is done. A heavier object also requires more work to be lifted.

How does the direction of the force affect the work done on an object being lifted?

The direction of the force applied to lift an object does not affect the work done on the object. As long as the force is in the same direction as the movement of the object, the work done is the same. However, if the force is applied in a direction perpendicular to the movement of the object, then no work is done on the object.

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