How Much Work Does Lifting and Pushing a Box Involve?

[jameo15]

A girl pushes a 3.0x10²N box along the floor for a distance of4.0m. The coeffifient ofkineticfriction is0.33. She then lifts the box straight up 1.2m. What is the total amount of work she has done?

I converted 3.0x10²N to kgs and found Ff. Then I found work. Then I found Wgrav.
Now I am stuck and do not know what to do to get the answer. ( 760J )

 

[Hootenanny]

A girl pushes a 3.0x10²N box along the floor for a distance of4.0m. The coeffifient ofkineticfriction is0.33. She then lifts the box straight up 1.2m. What is the total amount of work she has done?

I converted 3.0x10²N to kgs and found Ff. Then I found work. Then I found Wgrav.
Now I am stuck and do not know what to do to get the answer. ( 760J )

The total work done by the girl is simply the sum of the work done against friction and the work done against the weight of the box. Perhaps if you showed a little more detail of what you have done thus far, we could help you out a little more.

 

[jameo15]

d=4m
u=0.33
m=3.0x10²N=30.6kg
h=1.2m

Ff=uFn=0.33 x 30.6 x 9.8= 98.96N

Wgrav=-1mgh=-3(30.6)(9.8)(1.2)=359.9J

W=Fd=98.96 x 4
W=3879.2J

I do not believe this is right..

 

[Hootenanny]

d=4m
u=0.33
m=3.0x10²N=30.6kg
h=1.2m

Ff=uFn=0.33 x 30.6 x 9.8= 98.96N

Wgrav=-1mgh=-3(30.6)(9.8)(1.2)=359.9J

W=Fd=98.96 x 4
W=3879.2J

I do not believe this is right..

Why are you converting the weight to a mass and then back to a weight again? Note that the frictional force in this case may be written as:

F_r = \mu mg = \mu W

Equally, the work done against the weight of the box may be written:

\text{work} = mgh = W\cdot h

 

[jameo15]

Oh ok that makes more sense. Now I find Wnet correct?

Wnet= Wgrav ( answer from -1mgh formula) + Wp( answer from W=Fd formula)?

 

[Hootenanny]

Oh ok that makes more sense. Now I find Wnet correct?

Wnet= Wgrav ( answer from -1mgh formula) + Wp( answer from W=Fd formula)?

Correct, but please be aware that this quantity is positive since the girl is doing work.