A student hangs a weight on a newtonmeter. The energy now stored in the spring in the newtonmeter is 4.5times 10^-2 J The student then increases the weight on the newtonmete by 2.0 N Calculate the total extension of the spring. Spring constant=400N/m

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A student hangs a weight on a newtonmeter. The energy now stored in the spring in the newtonmeter is 4.5times 10^-2 J The student then increases the weight on the newtonmete by 2.0 N Calculate the total extension of the spring. Spring constant=400N/m

Answer 1

Let's begin calculating the total extension. Step 1: Calculate the original extension using the initial energy and the spring constant. Combine (generalizing variables): (1/2)kx^2 = E => x = sqrt(E/(0.5 * k)) Substituting the given values, we have: x = sqrt(4.5 * 10^-2J / (0.5*400 N/m)) = 0.015 m. So, 'x' i.e., the original extension of the spring when the weight was hanged is 0.015 metres. Step 2: Now given that the force F on the string increased, also increasing the energy by 2N (E’), The equation for energy E’ therefore, will be - E’ = E + 0.5Fx So, E’= 4.5 × 10^-2J + 0.5×2.0N×x= x^2×(0.5×400N/m) => x^2-(2/400)x-(4.5×10^-2 per 0.5) = 0 Inside the quadratic equation: {x = [-b ± sqrt(-b² - 4ac)] / 2a} take b (the y-intercept) to be -2/400, (-ve as directed to the left or because the standard formula for equation is ending up in negative); a = 1, and c = -4.5* (10^-2/0.5). Solving for x you find two values, but shall consider the positive one. In conclusion, original extension was 0.015 m. Additional force increased the extension of the spring which is our 'x'. Having applied the formula used to solve for different aspects physics formulae we have deep dived into this question and provided a precise solution. Let’s now enjoy tackling other questions -navigating through consideration, careful calculation and critical analysis to find optimal solutions as we learn.

Answer 2

The energy stored in a spring (otherwise known as elastic potential energy) can be computed using the equation E = (1/2)kx^2, where E is the energy, k is the spring constant, and x is the extension of the spring. In the question: Initial energy E is given as 4.5 x 10^(-2) J Spring constant k is 400 N/m We've been asked to figure out how the energy changes when an additional force F=2.0 N is applied to the spring. However, addition of this force to the spring means the energy will also increase. Since Energy E' = E + 0.5Fx, after substituting for the values in our question, we can then determine the extended distance x' where; E' = initial energy E + 0.5Fx = Work done by the force or potential energy stored in the spring.

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A student hangs a weight on a newtonmeter. The energy now stored in the spring in the newtonmeter is 4.5times 10^-2 J The student then increases the weight on the newtonmete by 2.0 N Calculate the total extension of the spring. Spring constant=400N/m

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