p=(1)/(2)kx^(2) As springs compress or stretch, potential energy in joules, p, is stored in them according to the given equation, where k is the spring constant in newtons per meter and x is the distance the springs compress or stretch. Tractor seats are mounted on springs to absorb impact. If the springs have a spring constant of 25,600 newtons per meter, what is the distance, in meters, the springs must compress or stretch in order to store 8 joules of potential energy? Choose 1 answer: (A) 0.000625 (B) 0.025 (C) 3200 (D) 102,400 meters

Question

p=(1)/(2)kx^(2) As springs compress or stretch, potential energy in joules,  p, is stored in them according to the given equation, where  k is the spring constant in newtons per meter and  x is the distance the springs compress or stretch. Tractor seats are mounted on springs to absorb impact. If the springs have a spring constant of 25,600 newtons per meter, what is the distance, in meters, the springs must compress or stretch in order to store 8 joules of potential energy? Choose 1 answer: (A) 0.000625 (B) 0.025 (C) 3200 (D) 102,400 meters

Bytelearn · Accepted Answer

Substitute values into equation: Plug in the given values for p and k into the equation p = (1/2)kx^2 to solve for x. p = 8 joules, k = 25,600 N/m. 8 = (1/2)(25,600)x^2.Eliminate fraction: Multiply both sides by 2 to get rid of the fraction. 2 * 8 = 25,600x^2. 16 = 25,600x^2.Solve for x^2: Divide both sides by 25,600 to solve for x^2. 16 / 25,600 = x^2. 0.000625 = x^2.Find final value of x: Take the square root of both sides to solve for x. x = √0.000625. x = 0.025 meters.

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