Question

A 2 kg brick of dimension 5 cm x 2.5 cm x 1.5 cm
is lying on the largest base. It is now made to
stand with length vertical, then the amount of
work done is (take, g = 10 ms)
(a) 35 J (0)5 J
(0) 7J (d) 9 J​

Answer 1

Given info : A 2 kg brick of dimension 5 m x 2.5 m x 1.5 m is lying on the largest base. It is now made to stand with length vertical.

To find : The amount of work done is ...

solution : as the brick is lying on the largest base, initial height of brick, h = 1.5 m

height from centre of mass of brick, h₁ = h/2 = 0.75m

so initial potential energy of brick, Ui = mgh₁

= 2kg × 10 m/s² × 0.75 m

= 15 J

now brick is now made to stand with length vertical.

so, final height of brick, H = 5 m

final height from centre of mass of brick, h₂ = 5/2 = 2.5 m

final potential energy of brick, Uf = mgh₂

= 2kg × 10 m/s² × 2.5 m

= 50 J

now work done = final potential energy - initial potential energy = 50 J - 15 J = 35 J

Therefore the amount of work done is 35 J.