A 6m*4m*2m block weighing 1500kg is resting with its 6m*4m file on the ground The work done in making - Physics - Work Energy And Power

Question

A 6m*4m*2m block weighing 1500kg is resting with its 6m*4m file on
the ground The work done in making it stand with its 2m*4m face on
the ground is

Shashwat Kumar · Answer

Consider that the entire block comprises of two blocks, I and II
whose dimensions you can see in the picture above Notice that in
bringing the entire system from the first configuration to the
second, as described in the question, block II does not change its
position at all The entire work done is done to pick up block I and
placing it on the top of block II In proper terms, total work done
= work done in raising block I from h=0 to h=2m (see the image)

Now, the block is of uniform density Mass is directly proportional
to volume MI + MII = 1500kg
volumeI = 4*4*2 = 32m3 and
volumeII = 2*2*4 = 16kg Hence, MI =
(32/48)*1500 = 1000kg

Hence work done = F d = MI*g*h = 1000*9 81*2 =
19,620J
Or, with g= 10m/s2, W = 1000*10*2 = 20,000J

A 6m*4m*2m block weighing 1500kg is resting with its 6m*4m file on the ground. The work done in making it stand with its 2m*4m face on the ground is

A 6m4m2m block weighing 1500kg is resting with its 6m4m file on the ground. The work done in making it stand with its 2m4m face on the ground is