An object of mass 1.6 kg is dropped from a height of 2.9 metres. It lands with a speed of 2.8...

Question:

An object of mass 1.6 kg is dropped from a height of 2.9 metres. It lands with a speed of 2.8 m/s.

How much energy was lost due to air resistance?

I Conservation of energy:

The conservation of energy before and after motion is conserved and is given by:

{eq}KE_{i} + PE_{i} = KE_{f} + PE_{f} + E_{loss} {/eq}

where:

  • {eq}KE {/eq} is the kinetic energy and is related to the velocity ({eq}v {/eq}) and mass ({eq}m {/eq}):

{eq}KE = \frac{1}{2}mv^{2} {/eq}

  • {eq}PE {/eq} is the potential energy and is related to the height ({eq}h {/eq}), mass ({eq}m {/eq}) and acceleration due to gravity ({eq}g {/eq}):

{eq}PE = mgh{/eq}

{eq}E_{loss} {/eq} is the energy lost from the motion and can be caused by parasitic, non-conservative forces like friction or air resistance.

Answer and Explanation:

Given:

  • {eq}m = 1.6\ kg {/eq}
  • {eq}d = 2.9\ m {/eq}

{eq}v = 2.8\ m/s {/eq}

We know from conservation of energy:

{eq}KE_{i} + PE_{i} = KE_{f} + PE_{f} + E_{loss} {/eq}

Plugging in the equations for {eq}PE {/eq} and {eq}KE {/eq}:

{eq}\frac{1}{2}mv_{i}^{2} +mgh_{i} =\frac{1}{2}mv_{f}^{2}+ mgh_{f} + E_{loss} {/eq}

Plugging in the values:

{eq}\frac{1}{2}\cdot 1.6 \cdot 0^{2} +1.6\cdot 9.8 \cdot 2.9 =\frac{1}{2}\cdot 2.9 \cdot 2.8^{2}+ 1.6\cdot 9.8 \cdot 0 + E_{loss} {/eq}

Solving for {eq}E_{loss} {/eq}:

{eq}E_{loss} = 41.6\ J {/eq}

The energy lost due to air resistance is 41.6 J.


Learn more about this topic:

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What is Energy Conservation? - Definition, Process & Examples

from ICSE Environmental Science: Study Guide & Syllabus

Chapter 1 / Lesson 6
96K

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