A body of mass five kilograms fell vertically from a height of 15 meters above the surface of the earth. Using the work–energy principle, find the kinetic energy of the body just before it hit the ground. Take 𝑔 equal to 9.8 meters per second squared.
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. 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. 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.
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. In order to calculate its velocity just before it hits the ground, we can use the equations of motion or SUVAT equations. We know that the initial speed is zero meters per second. The acceleration of the body is 9.8 meters per second squared. As it has fallen 15 meters, the displacement is 15. We can then calculate the value of 𝑣 using the equation 𝑣 squared is equal to 𝑢 squared plus two 𝑎𝑠. Substituting in our values, we have 𝑣 squared is equal to zero squared plus two multiplied by 9.8 multiplied by 15. The right-hand side simplifies to give us 294.
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. The kinetic energy of the body is therefore equal to a half multiplied by five multiplied by 294. This is equal to 735. The kinetic energy of the body just before it hit the ground is 735 joules.