A 9-kg mass is connected to a light spring (k = 36 N/m). The mass is pulled out

Question

A 9-kg mass is connected to a light spring (k = 36 N/m). The mass is pulled out 10 cm from its equilibrium position, and released at t = 0. How much work was done (by the applied force) as the mass was pulled out from 0 to x0? Hint: W = area under F(x) graph: What is the potential energy stored in the spring when it is stretched to x = 10 cm? [Hint: U=1/2kx2.] Once the mass is released, it begins oscillating in SHM. During the subsequent oscillations: What is its maximum kinetic energy? What is the system's mechanical energy (E = K +U)? The position of the mass is described by x(t) = xm cos(omega l): What are the values of xm and omega? What is its maximum speed?

Explanation / Answer

All of the work goes into potential energy. (1) W=(1/2)*k*(x0)^2 BY THE APPLIED FORCE. It is negative work by the spring if you look at it that way. (2) U=(1/2)*k*x^2=(1/2)*36*(0.1)^2=18*0.01=0.18J (3) Kmax=0.18J...as it passes the x=0 pt, all the elastic potential is kinetic energy. (4) The mechanical energy is 0.18J; @ Kmax, U=0 and at Umax, K=0. (5) xmax=0.1m because 0.1m is the maximum amplitude. w=sqrt(k/m) because if you take two derivatives of x(t) in order to find a(t)=-x0*w^2*cos(wt). At x0, cos(wt)=1 because the acceleration is a maximum (t=0). The force by the spring is -kx, so the acceleration (newton's second law) is -kx/m. You have -kx0/m=-x0*w^2. Thus w=sqrt(k/m). (6) vmax occurs at Kmax. v^2=0.18J*2/m...v=0.2 m/s

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