Magnus has reached the finals of a strength competition. In the first round, he has to pull a city bus as far as he can. One end of a rope is attached to the bus and the other is tied around Magnus's waist. If a force gauge placed halfway down the rope reads out a constant 2.500×10^3 Newtons while Magnus pulls the bus a distance of 2.312.31 meters, how much work does the tension force do on Magnus? The rope is perfectly horizontal during the pull.

Where F and d are the force and the total displacement. Note that in the definition the product is a scalar product since F and d are both vectors.

Take into account that according to third Newton's law the force that the rope does to Magnus is opposite to the force that Magnus does to the rope, therefore the scalar product will be negative due the rope's force goes against to Magnus displacement.

For calculating the work, we take 2500 N as the value for the force and 2.312 meters as the value for the displacement:

[tex]W=-2500 N * 2.312 m[/tex]

[tex]W=-5780 Nm = -5780 J[/tex]

hamzaahmeds hamzaahmeds · Answer 2 · 2021-10-17T16:38:19+02:00

This question involves the concept of work done and tension force.

The work done by tension force on Magnus is "-5775 J".

The work done by the tension force on Magnus is equal in magnitude but opposite in direction to the work done by Magnus to pull the city bus. This is due to the fact that tension force is a reaction force to the force applied by Magnus pulling the rope.

[tex]Work\ Done\ by\ Tension = W = - Work\ Done\ by\ Magnus\\W = - FdCos\theta[/tex]

where,

F = Force = 2500 N

d = displacement = 2.31 m

θ = angle between force and displacement = 0° (horizontal)

Therefore,

W = - (2500 N)(2.31 m)Cos 0°

W = - 5775 J

Learn more about work here:

https://brainly.com/question/13662169?referrer=searchResults

The attached picture explains the work done formula.