A 10 kg cart is attached to a rope that forces the cart to travel along a circular path. The rope has a length of 2.7 meters and has a rotational speed of 2.0 rotations per second.

Respuesta :

Answer:

  F = 4361.7 N

Explanation:

In this exercise the force of the string is asked, to solve it we use Newton's second law

          F - W = m a

force is the tension of the rope

         

the centripetal acceleration is

        a = v² / r

the angular and linear speed are related

        v = w r

let's substitute

          F - mg = m w² r

          F = m (g + w² r)

let's reduce the magnitudes to the SI system

       w = 2.0 rot / s (2π rad / 1 rev) = 4π  rad / s

let's calculate

       F = 10 (9.8 + (4π)² 2.7)

       F = 4361.7 N