Question

A 0.240 kg potato is tied to a string with length 1.90 m, and the other end of the string is tied to a rigid support. The potato is held straight out horizontally from the point of support, with the string pulled taut, and is then released.(a) What is the speed of the potato at the lowest point of its motion? (b) What is the tension in the string at this point?

Answer 1

Explanation:

Below is an attachment containing the solution.

Answer 2

Answer:

(a) $v=6.106m/s$

(b) $T=7.064N$

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

$mgr=1/2mv^2\\v^2=2gr\\v=\sqrt{2gr}$

Substitute the given

So

$v=\sqrt{2*(9.81m/s^2)(1.90m)} \\v=6.106m/s$

For part (b)

Consider forces on potato during circular motion

$T-mg=\frac{mv^2}{r}$

Substitute the given values to calculate tension in string at lowest point

$T=mg+\frac{mv^2}{r}\\ T=(0.240kg)(9.81m/s^2)+\frac{(0.240kg)(6.106m/s)^2}{1.90m}\\ T=7.064N$