Question

: A small block with mass 0.130 kg is attached to a string passing through a hole in a frictionless, horizontal surface. The block is originally revolving in a circle with a radius of 0.800 m about the hole with a tangential speed of 4.00 m/s. The string is then pulled slowly from below, shortening the radius of the circle in which the block revolves. The breaking strength of the string is 30.0 N. What is the radius of the circle when the string breaks

Answer 1

The radius of the circle when the string breaks is 0.0693 m.

Step-by-step explanation:

To find the radius of the circle when the string breaks, we need to consider the centripetal force acting on the block. The centripetal force is equal to the tension in the string.

Given:

• Mass of the block (m) = 0.130 kg
• Tangential speed (v) = 4.00 m/s
• Breaking strength of the string (F) = 30.0 N

The centripetal force is given by the equation:
F = mv^2/r

Substituting the given values:

30.0 N = 0.130 kg * (4.00 m/s)^2 / r

Solving for r:

r = 0.130 kg * (4.00 m/s)^2 / 30.0 N

r = 0.0693 m