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A body of mass five kilograms fell vertically from a height of 15 meters above the surface of the earth.
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Using the work–energy principle, find the kinetic energy of the body just before it hit the ground.
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Take 𝑔 equal to 9.8 meters per second squared.
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We begin by recalling that the work–energy principle states that the change in kinetic energy of an object is equal to the net work done on the object.
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We know that work is equal to force multiplied by distance, and the work–energy principle tells us that this is also equal to the change in kinetic energy.
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We also recall the formula for kinetic energy which is equal to a half 𝑚𝑣 squared, where 𝑚 is the mass of the object and 𝑣 is its velocity.
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In this question, we are told that a body of mass five kilograms falls vertically from a height of 15 meters above the surface of the earth.
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In order to calculate its velocity just before it hits the ground, we can use the equations of motion or SUVAT equations.
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We know that the initial speed is zero meters per second.
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The acceleration of the body is 9.8 meters per second squared.
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As it has fallen 15 meters, the displacement is 15.
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We can then calculate the value of 𝑣 using the equation 𝑣 squared is equal to 𝑢 squared plus two 𝑎𝑠.
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Substituting in our values, we have 𝑣 squared is equal to zero squared plus two multiplied by 9.8 multiplied by 15.
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The right-hand side simplifies to give us 294.
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Whilst we could square root both sides to calculate the value of 𝑣, we note that our kinetic energy formula contains the term 𝑣 squared.
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The kinetic energy of the body is therefore equal to a half multiplied by five multiplied by 294.
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This is equal to 735.
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The kinetic energy of the body just before it hit the ground is 735 joules.