Question

a rectangular block of dimensions 6m*4m*2m and of density 1.5 gm/c.c is lying on horizontal ground with the face of the largest area in contact with the ground.the work done in arranging it with it's smallest area in contact with the ground is (g=10ms^-2)​

Answer 1

Answer:

Explanation:

Volume of block=6*2*4=48m^3 AND MASS = v x D = 72 g also work done on block by gravity when we arrange in smallest area=MGH =10*72*2*1000 .=.1440KJ

Answer 2

Answer:

The work done is 1440 kJ.

Explanation:

Given that,

Density = 1.5 gm/c³

The volume of block is

$V = 6\times4\times2$

$V=48\ m^3$

We need to calculate the mass

Using formula of density

$\rho=\dfrac{m}{V}$

$m = \rho\times V$

Put the value into the formula

$m=1.5\times1000\times48$

The work done on block by gravity when we arrange in smallest are

We need to calculate the work done

$W= mgh$

Put the value into the formula

$W=1.5\times1000\times48\times10\times2$

$W=1440000\ J$

$W=1440\ kJ$

Hence, The work done is 1440 kJ.