Question

19. A rectangular block 5mx4mx2m lies on a table with its largest surface in contact with the table. The
work done to keep it so that the block rests on the smallest surface is , if its density is 600 kg/m
a) 352800 J
b) zero c) 376000J
d) 24,0000 J
​

Answer 1

Answer:

Answer is 376000J How means

Answer 2

Answer:

Explanation:

ANSWER

The volume of Rectangular block =5m×4m×2m

=40m

3

Mass = Volume × Density

=40m

3

×600km

−3

=24,000=2.4×10

4

hg

​

Potential energy of block when it is lying on its largest surface =mgh

=2.4×10

4

×9.8×1joule

Potential energy of the block when it is lying on

=mgh=2.4×10

4

×9.8×2.5

Work done = difference in potential energy

=(2.4×10

4

×9.8×2.5−2.4×10

4

×9.8×1)joule

=2.4×10

4

×9.8×1.5joule

=352800joule

​