Answer:
(a) [tex]v=6.106m/s[/tex]
(b) [tex]T=7.064N[/tex]
Explanation:
Given data
Mass m=0.240 kg
Length r=1.90 m
Required
(a) Speed v
(b) Tension T
Solution
For Part (a)
As the change in potential energy is equivalent to kinetic energy at lowest point.
So
P.E=K.E
[tex]mgr=1/2mv^2\\v^2=2gr\\v=\sqrt{2gr}[/tex]
Substitute the given
So
[tex]v=\sqrt{2*(9.81m/s^2)(1.90m)} \\v=6.106m/s[/tex]
For part (b)
Consider forces on potato during circular motion
[tex]T-mg=\frac{mv^2}{r}[/tex]
Substitute the given values to calculate tension in string at lowest point
[tex]T=mg+\frac{mv^2}{r}\\ T=(0.240kg)(9.81m/s^2)+\frac{(0.240kg)(6.106m/s)^2}{1.90m}\\ T=7.064N[/tex]
