Question

III
20. A uniform rod of length 2m and mass 5kg is
lying on a horizontal surface. The work done
in raising one end of the rod with the other
end in contact with the surface until the rod
makes an angle 30° with the horizontal is,
(g = 10ms-2)
1. 25J
2.50J
3.
$25 \sqrt{3}$
4.
$50 \sqrt{3}$


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Answer 1

Answer:

The correct answer is option $1.$ $25J$.

Explanation:

Concept:

By subtracting the difference between the original and final energies, the work is computed. We will therefore arrive at the answer for finishing the assignment by employing this.

Work done $=$ Final potential energy$-$Initial potential energy

Final potential energy minus work completed Initial potential energy is equal to zero, and the ultimate potential will be equal to, as can be seen.

Given:

length of rod $L=2m$

Mass of rod $m=5kg$

$g = 10ms^2$

To find:

We have to find the required work done.

Solution:

The initial potential energy is equal to zero, and the ultimate potential will be equal to,

$Final P.E.=mg*\frac{Lsin30}{2}$

When we solve it we got,

$Final P.E.=mg*\frac{L*\frac{1}{2}}{2}$

$Final P.E.=\frac{mgL}{4}$

Now, put the values in it,

$Final P.E.=\frac{5*10*2}{4}\\=25J$

The final potential energy is $25J$.

The formula of work done,

Work done $=$ Final potential energy$-$Initial potential energy

$=25-0\\=25J$

Thus, the correct answer is option $1.$ $25J$.

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