# A stone with a mass of 1.0 kg is tied to the end of a light string which keeps it moving in a circle

with a constant speed of 4.0 m/s on a perfectly smooth horizontal tabletop. The radius of the path is

0.60 m. How much work does the tension in the string do on the stone as it makes one-half of a

complete circle?

## NONE, there is no motion in the direction of the string tension. It continues at the same speed forever.

## 0j

## thanks

## To find the work done by the tension in the string, we need to use the equation:

Work = Force x Distance x cos(θ)

In this case, the force is the tension in the string, and the distance is the length of the arc that the stone travels along the circle. The angle (θ) is the angle between the direction of the force and the direction of the displacement.

Given that the stone travels half of a complete circle, the arc length is given by:

Arc Length = (1/2) x Circumference

Arc Length = (1/2) x 2πr

Arc Length = πr

Plugging in the values, the arc length becomes:

Arc Length = π x 0.60 m

Arc Length = 1.88 m

Now, let's calculate the tension in the string (force) using the centripetal force equation:

Centripetal Force = Mass x Acceleration

Where the acceleration is given by:

Acceleration = (Velocity^2) / Radius

Plugging in the known values:

Acceleration = (4.0 m/s)^2 / 0.60 m

Acceleration ≈ 26.67 m/s^2

Now, we can calculate the force:

Force = Mass x Acceleration

Force = 1.0 kg x 26.67 m/s^2

Force = 26.67 N

Lastly, we can calculate the work done by the tension in the string:

Work = Force x Distance x cos(θ)

Work = 26.67 N x 1.88 m x cos(180°)

Work = 26.67 N x 1.88 m x -1 (as cos(180°) = -1)

Work ≈ -100 J

Therefore, the tension in the string does approximately -100 Joules of work on the stone as it makes one-half of a complete circle. The negative sign indicates that the work done is against the direction of motion.