Question

A 0.400 kg potato is tied to a string with length 2.50 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. part a what is the speed of the potato at the lowest point of its motion? take free fall acceleration to be 9.80 m/s2 .

Answer 1

Let the following  be:v - speed at the lowest point, r - the radius, m - mass, g - acceleration due to gravity. 
Relating potential energy at the start to kinetic energy at the lowest point: mgr = mv^2 / 2 v^2 = 2gr ...(1) so we can say that:v = sqrt(2gr) = sqrt(2 * 9.81 * 2.50) = 7.00 m/s

Answer 2

Answer:

$v=7m/s$

Explanation:

The easiest way to solve this problem is using the law of conservation of energy which states that the energy in an isolated system remains constant.

We need to notice that the potato behaves the same as a pendulum.

We are going to set our frame of reference 2.5m bellow the rigid support, this way at the beginning the potato will be at its maximum high of 2.5m.

First, we need to remember that:

• gravitational energy is defined as $U=mgh$, where m is mass, g is gravity, and h is the height,
• kinetic energy is defined as $T=\frac{1}{2}mv^{2}$, where m is mass and v the speed.

Now using the law of conservation of energy and noticing that at the beginning we only have gravitational energy and at the end, we only have kinetic energy (because at the lowest point h=0) so, we get

$U=T$,

$mgh=\frac{1}{2}mv^{2}$,

$gh=\frac{1}{2}v^{2}$,

$v^{2}=2gh$,

$v=\sqrt{2gh}$,

$v=\sqrt{2(9.8)(2.5)}$,

$v=7m/s$.