A stone of mass 0.8 kg is attached to one end of a string. The string is 0.9 m long and is swirled in horizontal circular motion parallel to the ground. The string will break if the tension exceeds 60 N . Determine the maximum speed of the stone.

1. The centripetal force required to keep the stone moving in a circle is given by \[ F_c = m \frac{v^2}{r} \] where - \( m = 0.8\, \text{kg} \) is the mass of the stone, - \( r = 0.9\, \text{m} \) is the radius of the circle, and - \( v \) is the speed of the stone. 2. The maximum tension in the string is the same as the maximum centripetal force, i.e., \[ T_{\text{max}} = m \frac{v^2}{r} \] Given that \( T_{\text{max}} = 60\, \text{N} \), we can equate and solve for \( v \): \[ 60 = 0.8 \frac{v^2}{0.9} \] 3. Solve for \( v^2 \) by multiplying both sides of the equation by \( 0.9 \): \[ 60 \times 0.9 = 0.8 \, v^2 \] \[ 54 = 0.8 \, v^2 \] 4. Divide both sides by \( 0.8 \): \[ v^2 = \frac{54}{0.8} = 67.5 \] 5. Take the square root of both sides to find \( v \): \[ v = \sqrt{67.5} \approx 8.22\, \text{m/s} \] Thus, the maximum speed of the stone is approximately \( 8.22\, \text{m/s} \).