Question

A(n) 0.2 kg object is swung in a vertical circular path on a string 0.1 m long. The acceleration of gravity is 9.8 m/s2 . If a constant speed of 6.38 m/s is maintained, what is the tension in the string when the object is at the top of the circle.

Answer 1

Answer:

$T=83.37N$

Explanation:

Since the object is under a circular motion, according to Newton's second law, when the object is at the top of the circle we have:

$\sum F_y: T-mg=F_c$

Where $F_c$ is the centripetal force and is given by:

$F_c=ma_c=m\frac{v^2}{r}$

Replacing and solving for T:

$T=m\frac{v^2}{r}+mg\\T=0.2kg\frac{(6.38\frac{m}{s})^2}{0.1m}+0.2kg(9.8\frac{m}{s^2})\\T=83.37N$

Answer 2

The tension on the string when the object is at the top of the vertical circle is 79.45 N.

The given parameters;

• mass of the object, m = 0.2 kg
• radius of the circle, r = 0.1 m
• constant speed, v = 6.38 m/s
• acceleration due to gravity, g = 9.8 m/s²

The tension on the string when the object is at the top of the vertical circle is calculated as follows;

$T_{top} = \frac{mv^2}{r} - mg\\\\T_{top} = \frac{0.2 \times 6.38^2}{0.1} \ - \ 0.2(9.8)\\\\T_{top} = 81.41 - 1.96\\\\T_{top} = 79.45 \ N$

Thus, the tension on the string when the object is at the top of the vertical circle is 79.45 N.

Learn more about tension at the top of vertical circle: https://brainly.com/question/19904705