Question

A 6mx 4m x 2m block, weighing 1.5x10 kg is resting with the face of maximum area on the ground. The work done in making it stand with its face of minimum area on
the ground is (g=9.8m/s-)

(1) 44.1X10^3 J
(2) 14.7 x 10^3 J
(3) 29 4x10^3 J
(4) 39 2 x 10^3 J​

Answer 1

Figure not drawn to scale.

• Area of face will be maximum when the sides of the faces are 6 m and 4 m.See first figure.

• Area of face will be minimum when the sides of the faces are 4 m and 2 m.See second figure.

• The distance of C.O.M from ground in first case will be 1 cm.

• The distance of C.O.M from ground in second case will be 3 cm.

Concept

Work done = Force × displacement

• In this case ,force acting will be F = mg.

F = mg

= 1.5 × 10 × 9.8

= 14.7 × 10 N

Displacement = Distance the C.O.M moved

= 3 - 1

= 2 m

Hence, W = F.s

= 14.7 × 10 × 2

= 29.4 × 10 Joule.

Answer 2

Explanation:

Figure not drawn to scale.

Area of face will be maximum when the sides of the faces are 6 m and 4 m.See first figure.

Area of face will be minimum when the sides of the faces are 4 m and 2 m.See second figure.

The distance of C.O.M from ground in first case will be 1 cm.

The distance of C.O.M from ground in second case will be 3 cm.

Concept

Work done = Force × displacement

In this case ,force acting will be F = mg.

F = mg

= 1.5 × 10 × 9.8

= 14.7 × 10 N

Displacement = Distance the C.O.M moved

= 3 - 1

= 2 m

Hence, W = F.s

= 14.7 × 10 × 2

= 29.4 × 10 Joule.

hope this helps you!!