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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 :

Answer:

The tension  is:

[tex]T=125\: N[/tex]

Explanation:

We know that the centripetal force is:

[tex]F_{c}=ma_{c}[/tex]

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

[tex]T=F_{c}=ma_{c}[/tex]

a(c) is the centripetal acceleration.

[tex]T=F_{c}=m\frac{v^{2}}{R}[/tex]

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:

[tex]T=2\frac{10^{2}}{1.6}[/tex]

[tex]T=125\: N[/tex]

I hope it helps you!