Question

A 2-kg iron ball is swung in a horizontal

circular path at the end of a 1.6-m length of
rope. Assume the rope is very nearly hori.
zontal and the ball's speed is 10 m/s. Calcu-
late the tension in the rope.

Answer 1

The tension is:

`T=125\: N`

Step-by-step explanation:

We know that the centripetal force is:

`F_(c)=ma_(c)`

And the tension will be equal to the centripetal force at this point, so we have:

`T=F_(c)=ma_(c)`

a(c) is the centripetal acceleration.

`T=F_(c)=m(v^(2))/(R)`

v is the tangential speed (10 m/s)

R is the radius (1.6 m)

m is the mass of the ball (2 kg)

Therefore, the tension will be:

`T=2(10^(2))/(1.6)`

`T=125\: N`

I hope it helps you!