A woman with a mass of 63 kg is running at a speed of 4.8 m/s. What is her kinetic energy?
In this problem, you are given the speed of a person and asked for her kinetic energy. Kinetic energy is defined as the energy of motion, so you aren’t asked for new information in this problem but rather to restate what you know about motion in a slightly different way.
There is no need for a picture in most definition problems, and this is one of them. You are given velocity information at one point and asked for the closely-related kinetic energy information at that same point. A picture will not provide any additional insight or organization beyond what is already present in the problem.
In equation form, kinetic energy is defined as
KE = ½ mv^{2}
This is the only relation you need for this problem.
KE = ½ mv^{2}
KE = ½ (63 kg)(4.8 m/s)^{2}
KE = 730 J
There is no further calculation required in this problem.
Some books use K as a symbol for kinetic energy, and others use KE or K.E. These are all equivalent and refer to the same thing.
Only two significant figures were provided in this problem, so only two figures were kept in the solution.
The MKS unit of energy is the Joule. 1 J = 1 kg m^{2}/s^{2}. In this problem, mass was given in kg, and velocity^{2} is in (m/s)^{2}.
The MKS unit of energy is the Joule. 1 J = 1 kg m^{2}/s^{2}. In this problem, mass was given in kg, and velocity^{2} is in (m/s)^{2}.
Conservation of energy allows you to compare two (or more) different points in a motion and learn something about that motion through the comparison. In this case, you are given information only about one point in the motion, and are asked about that same point. Further, there is a very distinct pairing between speed and kinetic energy. If you move, you have kinetic energy associated with that motion, and the amount of energy you have depends on the speed at which you are moving. So to find KE from v or vice versa is merely a way of restating information that you already have.
KE = ½ mv^{2}
KE = ½ (63 kg)(4.8 m/s)^{2}
KE = 730 N
This problem is merely a definition problem. You will use the definition of kinetic energy in Conservation of Energy problems, much as you use F_{g} = mg as you solve force problems. Whenever you have information about the speed of an object, you can restate that information as its kinetic energy (or vice versa.)
The MKS unit of energy is the Joule. For context, one Joule is the amount of energy required to lift a one N (0.22 lb) object one meter (3.3 ft) in the air.