How much work is done by the lifting force?

[tica86]

An object of mass 0.550 kg is lifted from the floor to a height of 3.50m at a constant speed.
Can someone please check my answers, thanks!

1) How much work is done by the lifting force?

(0.550kg)(9.80m/s^2)(3.50)=18.9J

2) How much work is done by the Earth on the object?
I was told it was 0 but I think it's -18.9J
because W=-mgh

3) What is the net work done on the object?
Again I was told it was 18.9 J which I don't think it's correct
I think it's 0

4) What is the change in kinetic energy of the object?
0

5) What is the kinetic energy just before it hits the floor?
18.9J

6) What is its velocity?
This one I'm a little confused with, is it
(1/2)mv^2=18.9J
v^2=1/2m-18.9J


Also, I have no idea how to try and solve the following:
A force acts on an object of mass 0.425 kg. The force varies with position. Find the work done by the force in moving the object from 0.40m to 1.20m.

 

[PhanthomJay]

An object of mass 0.550 kg is lifted from the floor to a height of 3.50m at a constant speed.
Can someone please check my answers, thanks!

1) How much work is done by the lifting force?

(0.550kg)(9.80m/s^2)(3.50)=18.9J

2) How much work is done by the Earth on the object?
I was told it was 0 but I think it's -18.9J
because W=-mgh

3) What is the net work done on the object?
Again I was told it was 18.9 J which I don't think it's correct
I think it's 0

4) What is the change in kinetic energy of the object?
0

5) What is the kinetic energy just before it hits the floor?
18.9J

You went against the flow of opinion, and got them all correct, nice work! (I'm assuming that in (5) and (6) below, the object is dropped to the floor after its height of 3.5 m is reached).

6) What is its velocity?
This one I'm a little confused with, is it
(1/2)mv^2=18.9J

Yes

v^2=1/2m-18.9J

algebra error...v^2 = 2(18.9)/m, solve for v.

Also, I have no idea how to try and solve the following:
A force acts on an object of mass 0.425 kg. The force varies with position. Find the work done by the force in moving the object from 0.40m to 1.20m.

More info is needed to get a numerical value. Did you leave out something?

 

[tica86]

You went against the flow of opinion, and got them all correct, nice work! (I'm assuming that in (5) and (6) below, the object is dropped to the floor after its height of 3.5 m is reached).
Yes algebra error...v^2 = 2(18.9)/m, solve for v.More info is needed to get a numerical value. Did you leave out something?

Sorry I attached the image that deals with the last problem

 

[PhanthomJay]

Sorry I attached the image that deals with the last problem

If you can find the equation of that curve, then use the definition of work
,
W = \int Fdx evaluated betwen the 2 given distances. Or, as a close approximation, use the average force (the force at x = 0.8) times the distance traveled in that interval.

 

[issacnewton]

since analytical dependence of F(x) is not given, and since the plot F vs x is shown on graph paper, student probably is expected to find the area under the curve by counting unit squares to find the work