# 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.

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...

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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`

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