Centripetal force will be provided by the spring force. Let the elongation in the spring be x, then we can write
⇒kx=mω2(r+x)
⇒7.5x=2.5(0.2+x)
⇒x=50.5=0.1
Therefore, required tension will beT=kx=0.75N.
A small block of mass 100g is tied to a spring of spring constant 7.5Nm−1 and length 20cm. The other end of spring is fixed at a particular point A. If the block moves in a circular path on a smooth horizontal surface with constant angular velocity 5rads−1about point A, then tension in the spring is
Held on 6 Apr 2023 · Verified 6 Jul 2026.
0.75N
0.25N
0.50N
1.5N
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