Ten bricks each 6 cm thick and mass 1.5 kg, lie flat on a table. How much work is required to stick them one on the top of another. (Ans: 40 J)



Given:

Number of bricks = 10

Mass of each brick = m = 1.5 kg 

Height of each brick = 6 cm = 0.06 m 

Value of g = 9.8 m s⁻²


To Find:

Work done in sticking all the 9 brick one on the top of other  =  W = ?


Solution:

Let the 10th brick is laying on the table and we have to left the other 9 bricks one after other on the already laying 1oth brick.

As given the height of brick = h = 6 cm = 0.06 m so,

height of 1st brick = h
height of 2nd brick = 2h
height of 3rd brick = 3h
and so on

height of 9th brick = 9h

Work done in sticking all the 9 brick one on the top of other  = Sum of work done =  W = W₁ + W₂ + W₃ + W₄ + W₅ + W₆ + W₇ + W₈ + W₉ 

W = mgh + mg(2h) + mg(3h) + mg(4h) + mg(5h) + mg(6h) + mg(7h) + mg(8h) + mg(9h)

W = (1+2+3+4+5+6+7+8+9) mgh

W = (45) mgh

putting the corresponding values

W = 45ｘ1.5 kg ｘ 9.8 m s⁻² ï½˜ 0.06 m

W = 39.69 kg m² s⁻² 

or

W ≈ 40 J -------------Ans.




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