A 0.50-kg ball, attached to the end of a horizontal cord, is rotated in a circle of radius 1.9 m on a frictionless horizontal surface. If the cord will break when the tension in it exceeds 85 N, what is the maximum speed the ball can have?

Respuesta :

Answer:

[tex]\boxed {\boxed {\sf 18 \ m/s}}[/tex]

Explanation:

The ball is moving in a circle, so the force is centripetal.

One formula for calculating centripetal force is:

[tex]F_c= \frac{mv^2}r}[/tex]

The mass of the ball is 0.5 kilograms. The radius is 1.9 meters. The centripetal force is 85 Newtons or 85 kg*m/s².

  • [tex]F_c[/tex]= 85 kg*m/s²
  • m= 0.5 kg
  • r= 1.9 m

Substitute the values into the formula.

[tex]85 \ kg*m/s^2 = \frac{0.5 \ kg *v^2}{1.9 \ m}[/tex]

Isolate the variable v. First, multiply both sides by 1.9 meters.

[tex](1.9 \ m)(85 \ kg*m/s^2) = \frac{0.5 \ kg *v^2}{1.9 \ m}*1.9 \ m[/tex]

[tex](1.9 \ m)(85 \ kg*m/s^2) = {0.5 \ kg *v^2}[/tex]

[tex]161.5 \ kg*m^2/s^2 = 0.5 \ kg*v^2[/tex]

Divide both sides by 0.5 kilograms.

[tex]\frac {161.5 \ kg*m^2/s^2}{0.5 \ kg} = \frac{0.5 \ kg*v^2}{0.5 \ kg}[/tex]

[tex]\frac {161.5 \ kg*m^2/s^2}{0.5 \ kg} =v^2[/tex]

[tex]323 \ m^2/s^2 = v^2[/tex]

Take the square root of both sides of the equation.

[tex]\sqrt {323 \ m^2/s^2} =\sqrt{ v^2[/tex]

[tex]\sqrt {323 \ m^2/s^2} =v[/tex]

[tex]17.9722007556 \ m/s =v[/tex]

The original measurements have 2 significant figures, so our answer must have the same.

For the number we found, 2 sig fig is the ones place. The 9 in the tenth place tells us to round the 7 to an 8.

[tex]18 \ m/s =v[/tex]

The maximum speed is approximately 18 meters per second.