SOLUTION: A block of mass 98.0 g is sitting on a frictionless horizontal surface, and attached to a spring with spring constant 11.6 N/m. The block is then pulled a certain distance from its

Algebra ->  Test -> SOLUTION: A block of mass 98.0 g is sitting on a frictionless horizontal surface, and attached to a spring with spring constant 11.6 N/m. The block is then pulled a certain distance from its      Log On


   

Question 1130038: A block of mass 98.0 g is sitting on a frictionless horizontal surface, and attached to a spring with spring constant 11.6 N/m. The block is then pulled a certain distance from its equilibrium position and released such that it undergoes simple harmonic motion and attains a maximum acceleration of 1.8 m/s2. What is the maximum speed of the block as it oscillates in this way? Express your answer in m/s.
Answer by ikleyn(52748) About Me  (Show Source):
You can put this solution on YOUR website!
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Since the maximum acceleration is 1.8 m/s^2 (given), then the maximal force is

    F = ma (the Newton second law) = 0.098*1.8 = 0.1764 Newtons (N).

It is when the block is in his farthest displacement from the equilibrium position.



Since  F = k*L  (the spring law),  we can determine this maximum displacement

    L = F%2Fk = 0.1764%2F11.6 = 0.0152 meters (= 1.52 centimeters)


The potential energy of the block in this position is

    P = %28k%2AL%5E2%29%2F2.


Now apply the Mechanical Energy conservation law.


The kinetic energy (when the body has its maximum speed - at the equilibrium position) is


    E = %28m%2AV%5E2%29%2F2.


The conservation of energy law gives you  


    E = P,  which implies  %28m%2AV%5E2%29%2F2 = %28k%2AL%5E2%29%2F2,   or


    V = sqrt%28%28k%2AL%5E2%29%2Fm%29.


Substitute the known data to determine V:


    V = sqrt%28%2811.6%2A0.0152%5E2%29%2F0.098%29 = 0.165 m/s    ANSWER

Solved.