Calculating Work Done on a Wooden Block Resting on Surface PQYX

[thereddevils]

Homework Statement

A wooden block of mass 1.2 kg with dimensions 20cm x 10 cm x 10cm . It's resting on the surface PQYX (with base area 10cm x 10 cm , the vertical side would be 20cm in this case).What is the work done to enable it to rest on surface PQYX

Homework Equations

The Attempt at a Solution

On first thought , work done is simply the gain in potential energy ,

mgh=(1.2)(10)(0.2)=2.4 J

but that isn't the answer .

 

[tomwilliam]

In my understanding, there is no work done if the box is not moving.
You can say that the forces are balanced... that the frictional force acting up the slope is equal in magnitude to the component of the box's weight pointing down the slope, but as Work = (Force) x (distance in direction of the force), my understanding is that no work is done if the box remains steady.

 

[tiny-tim]

Hi thereddevils!

A wooden block of mass 1.2 kg with dimensions 20cm x 10 cm x 10cm . It's resting on the surface PQYX (with base area 10cm x 10 cm , the vertical side would be 20cm in this case).What is the work done to enable it to rest on surface PQYX

I don't understand …

is the block tilted? …

and where is the block moving from and to?

 

[thereddevils]

Hi thereddevils!


I don't understand …

is the block tilted? …

and where is the block moving from and to?

sorry , i should hv posted the complete question.

Figure 23 shows a wooden block of mass 1.2 kg with dimensions 20cm x 10 cm x 10cm . It's resting on the surface PQRS . What is the workdone to enable it to rest on the surface PQYX ?

PQRS is of base area (20 x 10) and PQXY (10 x10)

THe block is neither moving nor its tilted .

 

[tiny-tim]

Well, work done is force "dot" displacement …

if nothing has moved, then what's the displacement?