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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?

User Bfmags
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8.0k points

1 Answer

12 votes

Answer:


\boxed {\boxed {\sf 18 \ m/s}}

Step-by-step explanation:

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

One formula for calculating centripetal force is:


F_c= \frac{mv^2}r}

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.


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

Substitute the values into the formula.


85 \ kg*m/s^2 = (0.5 \ kg *v^2)/(1.9 \ m)

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


(1.9 \ m)(85 \ kg*m/s^2) = (0.5 \ kg *v^2)/(1.9 \ m)*1.9 \ m


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


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

Divide both sides by 0.5 kilograms.


\frac {161.5 \ kg*m^2/s^2}{0.5 \ kg} = (0.5 \ kg*v^2)/(0.5 \ kg)


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


323 \ m^2/s^2 = v^2

Take the square root of both sides of the equation.


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


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


17.9722007556 \ m/s =v

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.


18 \ m/s =v

The maximum speed is approximately 18 meters per second.

User Professor Falken
by
8.2k points
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