Question

A 1.50-kilogram ball is attached to the end of a 0.520-meter string and swung in a circle. The velocity of the ball is 9.78 m/s. What is the tension in the string? 74.6 N 276 N 28.2 N 3.39 N

Answer 1

Data:
Tension = Centripetal Force = ? (Newton)
m (mass) = 1.50 Kg
s (speed) = 9.78 m/s
R (radius) = 0.520 m (The piece of this rope with the ball tied in circular motion forms the circular radius).

Formula:
$F_{centripetal\:force} = \frac{m*s^2}{R}$

Solving:
$F_{centripetal\:force} = \frac{m*s^2}{R}$
$F_{centripetal\:force} = \frac{1.50*9.78^2}{0.520}$
$F_{centripetal\:force} = \frac{1.50*95.6484}{0.520}$
$F_{centripetal\:force} = \frac{143.4726}{0.520}$
$F_{centripetal\:force} = 275.9088... \to\:\boxed{\boxed{F_{centripetal\:force} \approx 276\:N}}\end{array}}\qquad\quad\checkmark$

Answer:
The tension in the string is 276 N

Answer 2

Answer:

The tension in the string is 276 N.

Explanation:

It is given that,

Mass of the ball, m = 1.5 kg

Length of the meter stick, r = 0.52 m

Velocity of the ball, v = 9.78 m/s

The tension acting in the string is balanced by the centripetal force acting on it. Its formula is given by :

$F_c=\dfrac{mv^2}{r}$

$F_c=\dfrac{1.5\ kg\times (9.78\ m/s)^2}{0.52\ m}$

$F_c=275.90\ N$

or

$F_c=276\ N$

So, the tension in the string is 276 N. Hence, this is the required solution.