# Suppose a rock of mass 2.2 [kg] is attached to a string of length 1.00 [m] is twirled around in a perfect circle as a speed of 2.5 [m/s]. Calculate the work ...

Suppose a rock of mass 2.2 [kg] is attached to a string of length 1.00 [m] is

twirled around in a perfect circle as a speed of 2.5 [m/s]. Calculate the work

done on the mass by the tension force exerted on the rock by the string

during one revolution of the rock. Ignore the effect of gravitational forces.

### 1 Answer | Add Yours

We need to deal with two aspects of this question: Work and Circular Motion.

- Work is equal to a force acting over a distance: W = Fd
- Circular motion is a lot like linear motion, but with the inclusion of the circle's radius, as well as the force keeping the body moving in a circle. This is the Centripetal Force, or `F_c` . The centripetal force always acts
**perpendicular to the linear motion.**This allows it to change the direction, but not the speed, of the object, and keep it moving in a circular path.

The solution is relatively simple. In order to find the Work done, we need to know what the force is, and the distance that it acted over. Again: W = Fd.

To find the distance traveled by the rock, we look at its path. It traveled in a circle; a circle's circumference is the "distance", or length, of its outer edge. Circumference is calculated by 2pi times the radius. For a radius of 1m;

2pi (1) = 6.28m

The force is acting over a distance of 6.28m

The force itself may be a bit less intuitive; it's the centripetal force. This force is the exact same thing as tension.

We know that centripetal force is equal to `(mv^2)/(r)`

So the force equals `((2.2)(2.5^2))/(1)`

Which works out to 13.75N

Work = F(d)

Work = 13.75 x 6.28

Work = 86.35 Joules

Since we're limited by 2 significant figures, `W = 86J`