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To elongate or to coil up a spring, some work is needed. This amount of work done is stored in the elongated or coiled up spring as its ELASTIC POTENTIAL ENERGY. The Spring releases this energy when it comes back to it’s original shape. The elongation or compression of a spring is proportional to the force applied on it. If a spring of length L1 is elongated to a length L2 then the extension (L2-L1) is proportional to the force applied on it. Let the applied force be f, then f ∞ (L2-L1) => f= k(L2-L1), Where k is called the force constant of the spring.

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Let’s calculate the average force acting on the spring When unstretched , the length of the spring becomes L1 & the force acting on it is 0, [ 0 because, when the spring comes back to its original shape, it releases its energy, i.e. work done is 0, hence force is also 0]. And when the length elongates to L2, the force becomes k(L2-L1). So, the average force acting on the spring is: ½[0+k(L2-L1) =½k(L2-L1) & L2-L1= displacement Now we know that formula of work done= force × displacement. So, work done on the spring will be = ½k(L2-L1) × (L2-L1) = ½k(L2-L1) 2

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So, the elastic potential energy (U) of the extended spring = Work done on the spring= ½k(L2-L1) 2 In the same way, the elastic potential energy of a compressed spring U= ½kx 2 where x= compression in length. So, basically the internal forces between the particles of the spring when it is stretched or compressed is called the elastic potential energy or the strain energy of the spring.

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Thank You…

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