Question

What tension must a 50.0 cm length of string support in order to whirl an attached 4,000.0 g stone in a circular path at 7.00 m/s?I

Answer 1

Answer:

The string must support the tension of 392 N.

Explanation:

The tension that the string must support should equal the centripetal force exerted on the on the stone as it goes in a circular path (because if the string supported less tension, it would break).

The centripetal force $F$ exerted on the stone is

$F=\frac{mv^2}{R}$

where

v = velocity of the stone in m/s

m = mass of the stone in kg

R = radius of the circular path.

Now the velocity of the stone is 7.00 m/s, the mass of the stone is 4000g or 4 kg (1000 g = 1kg), and the radius of the circular path is just the length of the string, and it is 50 cm or 0.5 m (100cm =1m); therefore, we get

m = 4kg

v =7m/s

R = 0.5m.

We put these values into the equation for the centripetal force and get:

$F=\frac{(4kg)(7.00m/s)^2}{0.5}$

$\boxed{F=392N }$

The centripetal force is 392 Newtons, and therefore, the tension that the string must support mus be 392 N.