Question

a rectangular plank of mass m1 and height a is on a horizontal surface. on the top of it another rectangular plank of mass m2 and height b is placed. the potential energy of the system is​

Answer 1

Answer:

P.E.

= (m1 + m2)g(Height of CM)

= (m1 + m2)g((a + b)÷2))

= (m1 + m2)(a + b)/2(Answer)

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Answer 2

The total potential energy of the given system  is  ( $\frac{m_{1}g }{2}$ + $m_{2}$ g )a +   $\frac{m_{2}gb }{2}$

Given:

A rectangular plank of mass m1 and height and another rectangular plank of mass m2 and height.

To Find:

The total potential energy of the given system

Solution:

Force on rectangular plank =       $m_{1}$*g

Force on rectangular plank placed on top of rectangular plank  = m2*g

Potential Energy = Force * Displacement

P1 =   $m_{1}$g*a/2 =    $\frac{m_{1}ga }{2}$

P2 = $m_{2}$ g * (a+ b/2)

=  $m_{2}$ ga +   $\frac{m_{2}gb }{2}$

Potential energy of system = p1 +p2

=    $\frac{m_{1}ga }{2}$  + $m_{2}$ ga +    $\frac{m_{2}gb }{2}$

= ( $\frac{m_{1}g }{2}$ + $m_{2}$ g )a +   $\frac{m_{2}gb }{2}$

Therefore,  The total potential energy of the given system  is  ( $\frac{m_{1}g }{2}$ + $m_{2}$ g )a +   $\frac{m_{2}gb }{2}$

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