The graph below is of the exerted by a function of the distance it is stretched

Question

The graph below is of the exerted by a function of the distance it is stretched from its unstretched length. Show calculations for each part below. 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 released. What is the kinetic energy of the cart just as the spring passes through its equilibrium length? 1. KE = 3J 2. KE = 0.42.J. 3. KE. = 1.5 J 4. KE = 0.21 J 5. No enough information What is the veiocity of the cart at this position? 1. V= 1.1m/s 2. V = 0.41 m/s 3. V = 2.2 m/s 4. V = 0.82 m/s 5. No enough information.

Explanation / Answer

Using the equation F = K x, we get...K = 4 / (1.5) N/ m KE at this instant = 1/2 Kx^2, since all spring elastic potential energy will be converted to Kinetic energy... so, KE = 0.5* (4 / 1.5) * (1.5^2) = 3 J Also, 0.5 m*v^2 = 3 J 0.5 * 5 * v^2 = 3J v = 1.095 m/s

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