Question

1. a 2-kg ball on a string is rotated in a circle of radius 10 m. if the string can withstand a maximum tension of 50 n, what is the maximum speed the ball can have without breaking the string?

Explanation:

Step1: Identify the centripetal force formula

The tension in the string provides the centripetal force for the circular motion of the ball. The formula for centripetal force is \( F_c = \frac{mv^2}{r} \), where \( F_c \) is the centripetal force (equal to the maximum tension here, \( T = 50\ N \)), \( m \) is the mass of the ball (\( m = 2\ kg \)), \( v \) is the speed of the ball, and \( r \) is the radius of the circular path (\( r = 10\ m \)).

Step2: Rearrange the formula to solve for \( v \)

Starting with \( F_c=\frac{mv^2}{r} \), we can rearrange it to solve for \( v \). Multiply both sides by \( r \): \( F_c\times r = mv^2 \). Then divide both sides by \( m \): \( \frac{F_c\times r}{m}=v^2 \). Take the square root of both sides: \( v = \sqrt{\frac{F_c\times r}{m}} \).

Step3: Substitute the known values into the formula

We know that \( F_c = T = 50\ N \), \( m = 2\ kg \), and \( r = 10\ m \). Substituting these values into the formula for \( v \), we get \( v=\sqrt{\frac{50\ N\times10\ m}{2\ kg}} \).

Step4: Calculate the value inside the square root first

First, calculate the numerator: \( 50\ N\times10\ m = 500\ N\cdot m \) (since \( 1\ N = 1\ kg\cdot m/s^2 \), this is also \( 500\ kg\cdot m^2/s^2 \)). Then divide by the mass: \( \frac{500\ kg\cdot m^2/s^2}{2\ kg}=250\ m^2/s^2 \).

Step5: Take the square root to find \( v \)

Now, take the square root of \( 250\ m^2/s^2 \). \( \sqrt{250}\approx15.81\ m/s \).

Answer:

The maximum speed the ball can have without breaking the string is approximately \( 15.8\ m/s \) (or more precisely \( \sqrt{250}\approx15.81\ m/s \)).