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A small block of mass 100 g is tied to a spring of spring constant 7.5 N m−1 and length 20 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 rad s−1about point A, then tension in the spring is

Question

A small block of mass 100 g is tied to a spring of spring constant 7.5 N m−1 and length 20 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 rad s−1about point A, then tension in the spring is

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Solution

The tension in the spring can be calculated using the formula for centripetal force, which is given by F = mω²r, where m is the mass of the block, ω is the angular velocity, and r is the radius of the circular path.

Step 1: Convert the mass of the block from grams to kilograms. 100 g = 0.1 kg

Step 2: Convert the length of the spring from cm to m. 20 cm = 0.2 m

Step 3: Substitute the values into the formula for centripetal force. F = (0.1 kg) * (5 rad/s)² * (0.2 m) = 0.5 N

The tension in the spring is equal to the centripetal force, so the tension is 0.5 N.

This problem has been solved

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