one end of a 1.0-m long string is fixed, the other end is attached to a 2.0-kg stone. the stone swings in a vertical circle, passing the top point at 4.0 m/s. the tension force of the string (in newtons) at this point is about?



Answer :

T = 51.6 N. At this point, the string's tension force (measured in newtons) is about vertical and passes the bottom point.

What is the string's tension?

A rope or rope is frequently pictured as having only one dimension, being massless, and having no cross section. The tension along the string is a constant and is equal to the strength of the forces exerted by the string's ends if there are no bend in the string, as there can be with vibrations or pulleys.

Explanation:

Since we are aware that the bottom location of a horizontal circular motion is where there is the greatest strain;

T − Mg = F-c = Mv²/r

Where F-c is centripetal force

Since T − Mg = Mv²/r

Let's make T the subject;

T = Mg + Mv²/r

Where g is the acceleration caused by gravity, which is 9.8 m/s2, T is the tension of the rope, M is the mass of the stone, v is its speed, and r is its length.

Considering the query,

M = 2Kg

v = 4 m/s

r = 1m

By entering the pertinent values, we may obtain T;

T = M(g + v²/r) = 2(9.8 + 4²/1)

T = 2(9.8 + 16)

T = 51.6N

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