Chapter 6.2, Problem 22SSC

To determine

The tension in the string.

Answer to Problem 22SSC

  $322.67\text{N}$

Explanation of Solution

Given:

The 40.0kg stone is whirled horizontally at a speed of $2.2\text{m/s}$ in a radius of 0.60m.

Formula used:

The centripetal force acting on an object is:

  $F=m{a}_c$

Where, $a_c=\frac{v^2}{r}$ is the centripetal acceleration, $v$ is the velocity of the object and r is the radius.

Calculation:

Now, as the stone is whirling in a circle, the tension in the string is getting balanced by the centripetal force.

  $T=F_c=m{a}_c=\frac{m{v}^2}{r}$

Using given values in above equation, solve:

  $\begin{array}{c}T=\frac{40{\left(2.2\right)}^2}{0.60}\\ =322.67\text{N}\end{array}$

Conclusion:

Thus, the tension in the string is $322.67\text{ N}$ .