Question

A meter scale of mass m initially vertical is dispalced at 45º keeping the upper and fixed, the change in P.E will be :-​

Answer 1

The change in P.E will be ΔPE= m×g×(√2−1)/2√2

Explanation:

The point marked on the ruler at a distance of L/2 is the centre of mass of the ruler.Now, this point is raised by a distance (L/2−h)

Change in potential energy of the whole body =ΔP.E= m×g× (L/2−h)

h/L/2 =cos 45º

h =  L/2 × cos 45

therefore, equation (1) becomes

ΔPE= m×g× (L/2−L/2×cos 45)

ΔPE= m×g×L/2(1−√2)

ΔPE= m×g×L/2 x √2√(1 - √2)

Now, since its a meters scale L = 1m

ΔPE= m×g×(√2−1)/2√2

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Answer 2

Hope the solution helps in clearing the doubt .

Thank you