# The mass of a cyclist together with his bike is 90 kg. Calculate the increase in kinetic energy...

## Question:

The mass of a cyclist together with his bike is 90 kg. Calculate the increase in kinetic energy as the speed increases from 6 km/h to 12 km/h.

## Change in kinetic energy:

The change in kinetic energy of a moving object is defined as the difference between its kinetic energies at the final and initial points. If an object does not change its position, then there will not be any change in its kinetic energy. And, it will also be equal to zero, when an object moves with uniform velocity or speed. Mathematically, the change in kinetic energy of an object is expressed as follows:

$$\Delta K.E = K.E_{f} - K.E_{i}$$

Where:

• {eq}\Delta K.E {/eq} is the change in the kinetic energy of the object.
• {eq}K.E_{i} {/eq} is the initial kinetic energy of the object.
• {eq}K.E_{f} {/eq} is the final kinetic energy of the object.

Identify the given information in the problem:

• Initial speed of the system of the cyclist together with his bike is {eq}v_{i} = 6.0 \, \rm km/h = 1.67 \, \rm m/s {/eq}
• Final speed of the system of the cyclist together with his bike is {eq}v_{f} = 12.0 \, \rm km/h = 3.33 \, \rm m/s {/eq}
• Mass of the system of the cyclist together with his bike is {eq}m = 90.0 \, \rm kg {/eq}

Change(increase) in kinetic energy of the system of the cyclist together with his bike can be expressed as follows:

{eq}\Delta K.E = K.E_{f} - K.E_{i} \\ \Rightarrow \Delta K.E = \dfrac{1 }{2 } m v_{f}^{2} - \dfrac{ 1}{2 } mv_{i}^{2} \\ \Rightarrow \Delta K.E = \dfrac{1 }{2 } m(v_{f}^{2} - v_{i}^{2} ) {/eq}

After plugging in the values, we have:

{eq}\Delta K.E = \dfrac{1 }{2 } (90.0 \, \rm kg) ( (3.33 \, \rm m/s)^{2} - (1.67 \, \rm m/s)^{2} ) {/eq}

Simplifying it further, we will get:

{eq}\color{blue}{\boxed{ \Delta K.E = 373.5 \, \rm J }} {/eq}