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 2 Joules, the spring has a stiffness of 200 Newton/meter, and the unstretched spring has a length of 50 centimeters. Find how much the spring is stretched from its natural length as the block moves in this circle.

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simarahluwaliasimar simarahluwaliasimar · Answer 1 · 2022-07-22T16:29:46+02:00

The spring is stretched about 3.5cm from its natural length as the block moves in this circle.

Acceleration is given by formula = [tex]\frac{v^{2} }{r}[/tex] where,..... (1)

v= velocity and r = total distance

In this case r = L+Δx where,

L is the length of unstretched spring

Δx = Change in Spring length

According to Hooke's law

F=kΔx …..(2) where,

F= force , k is constant and Δx= change in spring length

Also,

F=ma …..(3) where,

F= force , m= mass of body and a = acceleration

From equation (2) and (3)

ma = kΔx

Now, from equation (1)

mv²/L+Δx = kΔx

mv² = kΔx(L+Δx)…. (4)

Total energy(E) = 1/2mv² + 1/2kx²

From equation (4)

E = 1/2kΔx(L+Δx) + 1/2kΔx²

By solving this equation, this becomes

kΔx² + 1/2kΔxL - E = 0

Given,

k=200N/m L=50cm=0.5m E=2J

Δx = 0.035 m = 3.5cm

Hence, the spring is stretched about 3.5cm from its natural length as the block moves in this circle.

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