Question

Que. A uniform metre scale of mass 2 kg is suspended from one end. If it is displaced through an angle 60º from the vertical, the increase in its potential energy is (g = 9.8 m/s^2)..??

Pls give correct answer-​

Answer 1

Answer:

change in potential energy=mgh

m=2 kg

g=9.8 m/s^2

In calculating the potential energy, only the height of the centre of mass of the meter stick matters.

Initially the height of the center of mass of the meter stick can be considered be 0.

Finally, the height is, h= L/2 sin 60° where,

L=length of meter scale

Let L=1 m

h=12×3√2=3√4 m

change in potential energy=2×9.8×3√4=8.477 J

Explanation:

PLZ mark as BRAINLIEST

Answer 2

Answer:

Given:

Metre scale of mass 2 kg is suspended and displaced (rotated) by angle of 60°

To find:

Increase in potential energy

Calculation:

First of all , we need to draw the diagram of the rotation of the scale. And we need to remember that gravity acts at Centre of Mass of scale (i.e at half of scale length)

Using Trigonometry, we can say that the increase height of CENTER OF MASS of scale will be :

$\Delta l = \dfrac{l}{2} \{1 - \cos( \theta) \}$

So gain in potential energy will be :

$= m \times g \times \Delta l$

$= 2 \times 9.8 \times \dfrac{1}{2} \{1 - \cos( 60\degree) \}$

$= 9.8 \times (1 - \frac{1}{2} )$

$= \dfrac{9.8}{2}$

$= 4.9 \: joule$

So final answer is :

∆PE = 4.9 J