JEE MAIN - Physics (2023 - 6th April Morning Shift - No. 7)
A small block of mass $$100 \mathrm{~g}$$ is tied to a spring of spring constant $$7.5 \mathrm{~N} / \mathrm{m}$$ and length $$20 \mathrm{~cm}$$. 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 $$5 ~\mathrm{rad} / \mathrm{s}$$ about point $$\mathrm{A}$$, then tension in the spring is -
0.50 N
1.5 N
0.75 N
0.25 N
Explanation
In this problem, the spring is stretched due to the circular motion of the block, so the effective radius of the circular motion becomes the natural length of the spring plus the extension in the spring $(r=0.2+x)$.
The spring force, which is also the centripetal force, is given by $F_c=Kx=m\omega^2 r$, where $K$ is the spring constant, $x$ is the extension of the spring, $m$ is the mass of the block, $\omega$ is the angular velocity, and $r$ is the radius of the circular path.
Plugging in the values and solving for $x$ and $Kx$ gives the extension in the spring as $x = 0.1$ m and the tension in the spring (which is the spring force) as $Kx=7.5 \times 0.1 = 0.75$ N.
Substituting the given values:$7.5x = 0.1 \cdot (5^2) \cdot (0.2 + x),$
which simplifies to:
$7.5x = 5 \cdot (x + 0.2)$
Solving for $x$ gives $x = 0.1$ m. The tension in the spring is then $kx = 7.5 \cdot 0.1 = 0.75$ N.
So, the correct answer is $0.75$ N.
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