# A 0.600-kg particle has a speed of 2 m/s at point A and kinetic energy of 7.50 J at point B. What...

## Question:

A 0.600-kg particle has a speed of 2 m/s at point A and kinetic energy of 7.50 J at point B. What is (a) its kinetic energy at A? (b) its speed at B? (c) the total work done on the particle as it moves from A to B?

## Kinetic energy:

The kinetic energy is one of the form of energy. As we know that the energy is something that can not be created nor be destroyed but it will change into one another form by using different devices. The kinetic energy is the energy that is developed when the object is in motion. The kinetic energy is measured in Joule under the SI system.

Given data:

• Mass {eq}\left( m \right) = 0.6\;{\rm{kg}} {/eq}
• Speed {eq}\left( v \right) = 2\;{\rm{m/s}} {/eq}
• Kinetic energy {eq}\left( {K \cdot E} \right)_B = 7.5\;{\rm{J}} {/eq}

#### Part-a

Calculate the kinetic energy at {eq}A {/eq}

{eq}\left( {K \cdot E} \right)_A = \dfrac{1}{2}mv^2 = \dfrac{1}{2} \times 0.6 \times 2^2 = 1.2\;{\rm{J}} {/eq}

#### Part-b

Calculate the speed at {eq}B {/eq}

{eq}\begin{align*} \left( {K \cdot E} \right)_B &= \dfrac{1}{2}mv^2 \\ 7.5 &= \dfrac{1}{2} \times 0.6 \times v^2 \\ v &= 5\;{\rm{m/s}} \\ \end{align*} {/eq}

#### Part-c

Calculate the total work done {eq}\left( W \right) {/eq}

{eq}W = \left( {K \cdot E} \right)_B - \left( {K \cdot E} \right)_A = 7.5 - 1.2 = 6.3\;{\rm{J}} {/eq}

Hence, this is our required solution.