A block is attached to one end of a spring, while the other end of the spring is

Question

A block is attached to one end of a spring, while the other end of the spring is attached to a fixed axis. The block is moving in a horizontal circle centered at the axis at a constant speed, held in the circular path only by the spring. The system has a total energy of 1.50J, the spring has a stiffness of 200N/m, and the unstretched spring has a length of 40.0cm. Find how much the spring is stretched from its natural length as the block moves in this circle. ANY ALTERNATIVE SOLUTION WILL BE MUCH APPRECIATED.

FN l Ablock is attached to one end of a spring, while the other end of the spring is attached to a fixed axis. The block is moving in a horizontal circle centered at the axis at a constant speed, held in the circular path only by the spring. The system has a total energy of 1.50J, the spring has a stiffness of 200N/m, and the unstretched spring has a length of 40.0cm. Find how much the spring is stretched from its natural length as the block moves in this circle Mathematical Solution: l+ Ax kA.x l A -0.20mt, (0.20m)2 +4 mn +13cm Ax 3cm

Explanation / Answer

We know that,
Kinetic energy KE = 1/2 mv^2
and Potential Energy = 1/2kx^2

So, total Energy E = 1/2(mv^2 + kx^2)

Given,
E = 1.5 J
k = 200 N/m

Now,
we know that a = v^2 / l + x
and by F = ma = kx,
a = kx/m
So,
v^2 / l + x = kx/m
So,
v^2 = kx(l+x)/m

Hence
E = 1/2(mv^2 + kx^2)
=> E = 1/2(m[kx(l+x)/m] + kx^2)
=> E = 1/2(kxl+2kx^2)
=> E = 1/2kx(l+2x)
=> 3 = 200x(0.4+2x)
=> x = 0.032 m or 3 cm

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