Now let’s use the concept of elastic potential energy to solve a problem involvi

Question

Now let’s use the concept of elastic potential energy to solve a problem involving a spring. A glider with mass m=0.200kg sits on a frictionless horizontal air track, connected to a spring of negligible mass with force constant k=5.00N/m. You pull on the glider, stretching the spring 0.100 m, and then release it with no initial velocity. The glider begins to move back toward its equilibrium position (x=0). What is its speed when x=0.0800m?

What is the value of x when the object’s speed is 0.25 m/s?

Explanation / Answer

position x(t) = Asin(wt)

where A is amplitude = 0.100 m

w is frequency = [k/m]^0.5 = [5/0.2]^0.5 = 5 s^-1

speed v(t) = dx(t)/dt = wAcos(wt)

a) SO when x(t) = 0.080 m = 0.10*sin(wt)

=> wt = 53.13 degree

speed v(t) = 5*0.100*cos(53.13 degree) = 0.300 m/s

b) when v(t) = 0.25 m/s

0.25 = 5*0.100*cos(wt)

=> wt = 60 degree

SO position x(t) = 0.100*sin(60 degree) = 0.0866 m

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