A 0.40 kg ball, attached to the end of a horizontal cord, is rotated in a circle of radius 1.5 m on a frictionless horizontal surface. If the cord will break when the tension in it exceeds 63 N, what is the maximum speed the ball can have?

Respuesta :

Answer:

The maximum speed the ball can have is 15.37 m/s

Explanation:

Given;

mass of the ball, m = 0.4 kg

radius of the chord, r = 1.5 m

maximum tension on the chord, T = 63 N

The maximum tension on the chord is given by;

[tex]T_{max} = \frac{mv_{max}^2}{r} \\\\v_{max}^2 = \frac{T_{max} *r}{m}\\\\ v_{max} = \sqrt{\frac{T_{max} *r}{m}} \\\\ v_{max} =\sqrt{\frac{63 *1.5}{0.4}}\\\\v_{max} = 15.37 \ m/s[/tex]

Therefore, the maximum speed the ball can have is 15.37 m/s