2 kg mass is attached to the end of spring and pulled back 90 degree before bein

Question

2 kg mass is attached to the end of spring and pulled back 90 degree before being let go as shown below (imagine a pendulum). The spring has a spring constant of 100 N/m and is initially at its natural length of 1 meter before the mass is released. When the mass reaches the bottom of its swing, the spring will have stretched. Write a series of equations that could used to determine how far the spring has stretched and also the speed of the mass at the bottom of its swing. Solve your equations to find how far the spring stretches and the speed of the 2 kg mass when it is at the bottom of its swing.

Explanation / Answer

When the spring is horizontal and the mass at a height we take that as the 0 reference point

Potential energy U =0  

Kienetic energy K =0 as the system is at rest and spring is has its natural length 1m

When the mass is at the bottom, let the spring is stretched bya length of x

Potential energy of the Spring   Us = 0.5 kx2

potential energy of the mass = -mg(l+x) ,

KE of the mass K = 0.5 mv2

conserving the total energy

-mg(l+x) +0.5kx2 +0.5 mv2 = 0

mv2 = 2mg(l+x) - kx2   ---(1)

The mass is moving perpendicular to the spring length when it is at its bottom

The spring force = -kx

centrifugal force on the mass = mv2/(l+x)

gravitational force   = mg

balancing the force in the vertical direction

-kx +mg+mv2/(l+x) =0

mv2 = (l+ x) (kx-mg)   ---- (2)

eliminate v from 1 and 2

(l+ x) (kx-mg)   = 2mg(l+x) - kx2  

(1+x)( 100x - 2*9.8) = 2*2*9.8 (1+x) -100x2

solving for x wwe get

x= 0.4489 m = 44.89 cm

mv2 = (l+ x) (kx-mg)  

v2 = (1+0.4489)(100*0.4489 -2*9.8) /2

v= 4.28 m/s

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