Up Next. What is the PE of the spring? This is also called as the energy density of the strained wire. k The total elastic energy placed into the spring from zero displacement to final length L is thus the integral. The work done = energy stored in stretched string = F.dx. This is the currently selected item. Khan Academy is a 501(c)(3) nonprofit organization. Elastic Potential Energy Formula Questions: 1) You have an elastic spring that has a spring constant of 1.5 x 10-2 Newtons per meter, and the spring is compressed by 15.0 cm. Gravitational potential energy and conservative forces. Next lesson. This is an expression for strain energy or potential energy per unit volume of stretched wire. The elastic potential energy stored by the spring when it is has been stretched 0.40 m is 0.60 J. This equation is derived from Hooke’s Law, which states that force exerted to stretch a spring is directly proportional to its displacement. Different Forms of Expression of Strain Energy per Unit Volume: By definition of Young’s modulus of elasticity 2) A spring with a spring constant k = 800 N/m has been compressed, and 196 J of potential energy is stored. derive an expression for elastic potential energy in a stretched wire and show that elastic energy . derivation of elastic potential energy. Spring potential energy and Hooke's law review. Practice: Calculating elastic potential energy. U = 0.60 J. Calculating spring force . unit is J m-3 and its dimensions are [L-1 M 1 T -2]. Experimentally, we find. PE or U = is the potential energy of the object m = refers to the mass of the object in kilogram (kg) g = is the gravitational force $$m s^2$$ h = height of the object in meter (m) Besides, the unit of measure for potential energy is Joule (J). U = 1/2(7.50 N/m)(0.40 m) 2. The elastic potential energy formula is P.E = 1/2 kx2. This is known as Hooke's law. Expert Answer: Hooke's Law, F = -kx, where F is the force and x is the elongation. ... and l The elastic potential energy equation is used in calculations of positions of mechanical equilibrium. ... Lastly, throughout the derivation for stretching the body, signs are ignored, If I understand correctly. https://www.toppr.com/guides/physics-formulas/elastic-potential-energy-formula Elastic Potential Energy of a Strained Body (A) Using ## Y = \frac {stress}{strain}## we get ##F = \frac {AY}{L} * x## where ##F## is the restoring force, ##x## is the distance the body is stretched by. We might write this in equation form as F = k x. In the equation above, "P.E" is the elastic potential energy and expressed in joules, "X" is the displacement and "K" is the spring constant. How much work is done when we stretch a spring a distance x from its equilibrium position? Derivation of the Formula. The spring constant k = 1.5 x 10-2 Newtons/m and the Δs = 15.0 cm = 0.15 m. 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