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The total energy of a mass–spring system is the sum of its kinetic and potential energy:
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- Given that the Spring energy is ½ kx2, where x is the displacement of the spring from its equilibrium position, and k is the spring constant, what are the units of k, if the spring energy is in Joules (kg m2/s2)?arrow_forwardAt the park you see a child swinging. You notice it takes about 4.1 seconds for the child to make one complete swing (to return to their starting point). What is the period of the swing? (Report to 2 decimal places, and include the symbol for the unit of measure.For example, if the unit is meters, the symbol is m.)arrow_forwardA mass attached to a spring has non-constant acceleration a(t) = - cos ((pi*t)/4) * m / (s ^ 2) where t is time measured in seconds. If the initial velocity is zero, v\{0\} = 0m / s then what is the net change in position from t = 0 and t = 3 s? Give a numeric answer rounded to two decimal places.arrow_forward
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- If you did the previous question right, you hopefully got an expression for yo. You may notice that you can simplify the differential equation a little bit: d'y k (y – yo) dt2 т The parameter yo now plays the roll of the "relaxed length". A better term may be "equilibrium value for y". But mathematically, it's identical to a relaxed length with the spring as the only force. We continue using this equation: y(t) = Y0 + A cos(wt + y) Now, solve for A (in cm) with these parameters. Again, if you need more information, enter -100000. The parameters are: •m = 200 grams • Yo = (equilibrium value) = 40 cm • k = (spring constant) = 0.03 N/cmarrow_forwardA mass of 18 kilograms stretches the spring 0.3 meters. Use this information to find the spring constant (use g = 9.81 m/s² as the acceleration of gravity). k = 588 The previous mass is detached from the spring and a mass of 14 kilograms is attached. This mass is displaced 0.7 meters below equilibrium and then launched with an initial velocity of 1 meters/second. Find the equation of motion. Note: When solving this problem, consider positions below equilibrium to be positive. x(t) =arrow_forwardCan you help me solve these equations?arrow_forward
- Principles of Physics: A Calculus-Based TextPhysicsISBN:9781133104261Author:Raymond A. Serway, John W. JewettPublisher:Cengage Learning