Conservation of energy of an ice flake

[lauriecherie]

Homework Statement

A 1.90 g ice flake is released from the ege of a hemispherical bwon whose radius r is 28.0 cm. The flake bowl contact is frictionless. What is the speed of the flake when it reaches the bottom of the bowl.

Homework Equations

I took KINETIC ENERGY = GRAVITATIONAL POTENTION ENERGY. since KINETIC ENERGY = .5*M*V^2, and GRAVITATIONAL POTENTIAL ENERGY = MGH, i canclled the masses and solved for V. That gave me VELOCITY = SQUARE ROOT OF (2*G*R).

The Attempt at a Solution

I got 23.44 for my answer but after looking over my calculations I noticed that the radius is given in cm. Need I convert it to meters? If I do my answer comes out as 2.344 m/s. Is this correct now?

 

[lauriecherie]

Nevermind. It is correct.

 

[nlightner]

I would like to clarify that the conservation of energy principle states that energy cannot be created or destroyed, but can only be transferred or converted from one form to another. In the case of the ice flake, it starts with potential energy due to its position at the edge of the bowl and as it falls, this energy is converted into kinetic energy. Therefore, the total energy (potential + kinetic) remains constant.

To answer the question, we can use the conservation of energy equation: Potential Energy = Kinetic Energy. In this case, the potential energy is due to the gravitational force, which can be calculated using the formula PE = mgh, where m is the mass of the ice flake, g is the acceleration due to gravity, and h is the height from which the flake is released (in this case, the radius of the bowl).

Therefore, the potential energy at the edge of the bowl is PE = (1.90 g)(9.8 m/s^2)(0.28 m) = 5.2928 x 10^-2 J. This energy is then converted into kinetic energy at the bottom of the bowl, which can be calculated using the formula KE = 1/2 mv^2, where m is the mass of the ice flake and v is its velocity at the bottom of the bowl.

Setting the two energies equal to each other, we get:

PE = KE
5.2928 x 10^-2 J = 1/2 (1.90 g) v^2
v = √(2(5.2928 x 10^-2 J)/(1.90 g))
v = 2.344 m/s

Therefore, the speed of the ice flake when it reaches the bottom of the bowl is 2.344 m/s. It is important to note that the units for all the values should be consistent, so converting the radius to meters is necessary in order to get the correct answer. This is because the units must cancel out in order to get the correct unit for velocity (m/s).

In conclusion, the conservation of energy principle can be used to determine the speed of the ice flake at the bottom of the bowl by equating the potential energy at the edge of the bowl with the kinetic energy at the bottom of the bowl. The resulting velocity is 2.344 m/s, which is indeed the correct answer.