47. 100.0 J of work accelerates a 0.500 kg hockey puck across an ice rink (a frictionless environment). The total kinetic energy gained by the puck is?
a. 0.500 J b. 20.0 J c. 50.0 J d. 100.0 J

d. 100.0 J

Explanation:

To solve this problem we must use the theorem of work and energy conservation. This tells us that the mechanical energy in the final state is equal to the mechanical energy in the initial state plus the work done on a body. In this way we come to the following equation:

E₁ + W₁₋₂ = E₂

where:

E₁ = mechanical energy at state 1. [J] (units of Joules)

E₂ = mechanical energy at state 2. [J]

W₁₋₂ = work done from 1 to 2 [J]

We have to remember that mechanical energy is defined as the sum of potential energy plus kinetic energy.

The energy in the initial state is zero, since there is no movement of the hockey puck before imparting force. E₁ = 0.

The Work on the hockey puck is equal to:

W₁₋₂ = 100 [J]

100 = E₂

Since the ice rink  is horizontal there is no potential energy, there is only kinetic energy

Ek = 100 [J]

It can be said that the work applied on the hockey puck turns into kinetic energy.

i don't understand is there any other way you can rephrase it plz let me know in the comments and i will do it.

explanation:

the inner core.

good luck: )

klicker