a stone with a mass of 0.700 kg is attached to one end of a string 0.700 m long. the string will break if…
a stone with a mass of 0.700 kg is attached to one end of a string 0.700 m long. the string will break if its tension exceeds 55.0 n. the stone is whirled in a horizontal circle on a frictionless tabletop; the other end of the string remains fixed. for related problemsolving tips and strategies, you may want to view a video tutor solution of force in uniform circular motion. part a find the maximum speed the stone can attain without breaking the string. express your answer in meters per second. v_max = m/s
Answer
Explanation:
Step1: Identify centripetal - force formula
The centripetal force $F_c$ acting on the stone in circular motion is provided by the tension $T$ in the string. The formula for centripetal force is $F_c=\frac{mv^{2}}{r}$, where $m$ is the mass of the stone, $v$ is its speed, and $r$ is the radius of the circular path. Here, $F_c = T$, $m = 0.700$ kg, $r=0.700$ m, and the maximum tension $T_{max}=55.0$ N.
Step2: Solve for maximum speed
We have $T_{max}=\frac{mv_{max}^{2}}{r}$. Rearranging the formula for $v_{max}$, we get $v_{max}=\sqrt{\frac{T_{max}r}{m}}$. Substitute $T_{max} = 55.0$ N, $r = 0.700$ m, and $m = 0.700$ kg into the formula: [v_{max}=\sqrt{\frac{55.0\times0.700}{0.700}}] [v_{max}=\sqrt{55.0}] [v_{max}\approx7.42] m/s
Answer:
$7.42$ m/s