A system consists of a spring with force constand k=1050 N/m, length L = 1.60 m,

Question

A system consists of a spring with force constand k=1050 N/m, length L = 1.60 m, and an object of mass m=5.2kg attached to the end. The object is placed at the level of the point of attachment with the spwring unstretched (held perfectly horizontally) and the it is released so that it is swings llike a pendulum. Find the y position of the object at the lowest point.

I know that at the lowest point the spring will be stretched, so the lowest point will be more than 1.5 m below the point of attachment, but I cannot get the right answer. I tried to use the conservation of energy equations, and also a force balance, but I cannot get the right answer, which is that the spring will be stretched 0.135m, so it is 5.335m directly below the point of attachment at its lowest point. I have spent HOURS trying to get the answer. PLEASE HELP!

Explanation / Answer

Using conservation of energy at lowest point (let it be r from the point of attachment).and initial point(

Then m*r*g=1/2[ 1050*(r-1.6)^2+ m*v^2]

and using force equation at the lowest point: K*(r-1.6)=m*g+m*v^2/r gives v^2= [K*(r-1.6)-m*g]*r/m

Putting in equation above: 2*m*r*g=1050*(r-1.6)^2+[K*(r-1.6)-m*g]*r

Which forms the quadratic equation: 102.024r=1050(r^2-3.2r+2.56)+1050r^2-1680r-51.012r

Quadratic equation: 2100r^2-5193.036r+2688=0

Gives r=1.735212451 meters: and extension(y)= 0.135212451 meters

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