Question

you may complete this review up to 3 times. you can earn up to 10 points of extra credit on the unit 4 quiz.
round all numerical answers to the fourth digit behind the decimal place. enter all numerical answers in standar
floating notation (normal decimals), not scientific notation.

question 5 1 pts
a string can withstand a maximum tension force of 243 newtons. the string has a length of 1
meter. a 5 kg object is attached to the end of the string, and it is twirled over a students head
in a horizontal circular path. what is the maximum speed, in meters per second, that the
object can be rotated before the string breaks?

Explanation:

Step1: Identify the centripetal force formula

The centripetal force \(F_c\) is given by \(F_c=\frac{mv^{2}}{r}\), where \(m\) is the mass of the object, \(v\) is the speed, and \(r\) is the radius of the circular path. In this case, the tension in the string provides the centripetal force (\(F = F_c\)).

Step2: Rearrange the formula to solve for \(v\)

Starting with \(F=\frac{mv^{2}}{r}\), we can solve for \(v\). First, multiply both sides by \(r\): \(Fr = mv^{2}\). Then divide both sides by \(m\): \(v^{2}=\frac{Fr}{m}\). Finally, take the square - root of both sides: \(v=\sqrt{\frac{Fr}{m}}\)

Step3: Substitute the given values

We are given that \(F = 243\space N\), \(m = 5\space kg\), and \(r = 1\space m\). Substituting these values into the formula \(v=\sqrt{\frac{Fr}{m}}\), we get \(v=\sqrt{\frac{243\times1}{5}}\)

Step4: Calculate the value of \(v\)

First, calculate \(\frac{243}{5}=48.6\). Then, \(\sqrt{48.6}\approx6.9714\)

Answer:

\(6.9714\)