A boy swings a ball on a string at constant speed in a horizontal circle that has a circumference equal to 6 m. What is the work done on the ball by the 10-n tension force in the string during one revolution of the ball?

In this problem, the force is the tension in the string, while the object is the ball. The tension is always radial (towards the centre of the circle), while the ball moves tangentially to the circle: this means that the tension and the displacement are always perpendicular to each other, so

[tex]\theta=90^{\circ}, cos \theta = 0[/tex]

and so the work done is zero.

whitneytr12 whitneytr12 · Answer 2 · 2021-10-15T20:18:48+02:00

The work done on the ball swung on a string in a horizontal circle due to the tension force of 10 N in the string during one revolution of the ball is zero.

The work done by the tension force on the ball is given by:

[tex] W = F*d*cos(\theta) [/tex] (1)

Where:

F: is the tension force = 10 N

d: is the displacement

θ: is the angle between the force and the displacement

The displacement is given by the circular path of circumference 6 meters.

In one point of the circular path, the direction of the tension force is to the center of the circumference and the displacement is orthogonal to this direction, so the angle between them is 90° (see the picture below).

The work is then (eq 1):

[tex] W = F*d*cos(\theta) = 10 N*6 m*cos(90) = 0 [/tex]

Therefore, the work done on the ball by the tension force is 0 (zero).

Find more about work here:

https://brainly.com/question/2426583?referrer=searchResults

https://brainly.com/question/13852139?referrer=searchResults

I hope it helps you!