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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.

User Wengseng
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1 Answer

1 vote

Answer:

F = 4361.7 N

Step-by-step 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

User Luis Sep
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