Question

A uniform rectangular marble slab is 3.4m long and 2.0m wide. It has a mass of $200\,kg$. If it is originally lying on the flat ground, how much work is needed to stand it on one vertex $?$$\left( {{\text{Assume slab position is changed slowly}}} \right)$
A. $2.0\,KJ$
B. $3.0\,KJ$
C. $3.35\,KJ$
D. $3000\,KJ$

Answer 1

Hint: In the question, length, width and mass of the rectangular marble slab is given. By using the equation of work we get the value of the work done needed to stand the slab on the one vertex.
Formula used
The expression for finding the work need to stand it on one vertex is
$W = mg \times h$
Where,
$m$ be the mass of the rectangular marble slab, $g$be the acceleration due to the gravity and $h$ be the height of the slab.

Complete step by step solution:
Given that,
Length of the slab $l$=$3.4\,m.$
Width of the slab $b = 2.0\,m.$
Mass of the rectangular slab $m = 200\,kg.$
When the rectangular marble slab on shorter end, the height of center of mass of shorter side will be,
$h = \dfrac{b}{2}$
Substitute the known values in the above equation, we get
$h = 1\,m.$
Now, we have to find the work done needed to stand the rectangular marble slab is,
$W = F \times g$
Substitute the known parameters in the above equation, we get
$W = mg \times h$
Substitute the known values in the above equation, we get
$
  W = 200 \times 9.81 \times 1 \\
  W = 1962\,J. \\
$
Therefore, the work done needed to stand the rectangular marble slab on one vertex is $1962\,J\,\left( {{\text{Approx}}} \right)$.

Hence from the above options, option A is correct.

Note: In the question, we know that some external force acts on the body for moving it to some distance. So we know the work done becomes change. But here the force is applied to the center of mass of the object, so the minimum work has to be done for the lifting purposes. So here we include the height of the slab in the equation, then we get the value of the work done.