The graph on the right is of the force exerted by a spring as a function of the

Question

The graph on the right is of the force exerted by a spring as a function of the distance it is stretched from its unstretched length. Show calculations for each part below. What is the spring constant of this spring? What is the elastic potential energy if the spring is stretched 1.5 m from its equilibrium length? A low-friction cart of mass 5.0 kg is attached to the spring, the spring is stretched 1.5 m from its equilibrium length, and then the cart is released. What is the kinetic energy of the cart just as the spring passes through its equilibrium length? What is the velocity of the cart at this position?

Explanation / Answer

a) as shown in graph k is slope so k= (4-0)/(1.5-0) = 2.67N/m b) Elastic potential enrgy = kx^2/2 = 2.67 *2.25 /2 = 3.00375 j c)at equilubrium E = KE so KE = 3.00375 J mv^2/2 = 3.00375 J v= sqrt(6/5) = 1.095 m/s plz rate me..............................

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