Question Video: Finding the Kinetic Energy of a Body Projected Downward When It Is about to Hit the Ground | Nagwa Question Video: Finding the Kinetic Energy of a Body Projected Downward When It Is about to Hit the Ground | Nagwa

# Question Video: Finding the Kinetic Energy of a Body Projected Downward When It Is about to Hit the Ground Mathematics

A body of mass 400 g was projected at 4 m/s vertically downward from a point 5 m above the ground. Use the work–energy principle to calculate the body’s kinetic energy when it was about to hit the ground. Take 𝑔 = 9.8 m/s².

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### Video Transcript

A body of mass 400 grams was projected at four meters per second vertically downward from a point five meters above the ground. Use the work–energy principle to calculate the body’s kinetic energy when it was about to hit the ground. Take 𝑔 equals 9.8 meters per second squared.

We can draw a diagram with the labeled information from the problem about our object. It has a mass of 400 grams. It’s thrown vertically downward at a speed of four meters per second from a height of five meters. And we are looking to find the kinetic energy when the object is just about to hit the ground. We are instructed to use the work–energy principle. In equation form, this means that the net work done on an object is equal to the change in kinetic energy of the object. The net work done on an object is equal to the net force on the object times the displacement. So we can replace 𝑊 with 𝐹 times 𝑑. Change in kinetic energy means the final kinetic energy minus the initial kinetic energy. We should recall that the kinetic energy of an object is equal to one-half 𝑚𝑣 squared, where 𝑚 is the mass of the object measured in kilograms and 𝑣 is the speed of the object measured in meters per second.

As we’re solving for the final kinetic energy, we don’t need to change this variable. However, the problem did not give us the initial kinetic energy, so we need to replace this variable with an equal form, which would be one-half 𝑚𝑣 initial squared. The net force 𝐹 acting on the object is the force due to gravity as it’s pulling the object to the ground. We should remember that the force due to gravity is the mass of the object measured in kilograms times the acceleration due to gravity, which was given to us as 9.8 meters per second squared in the problem. We can go ahead and replace the net force in our equation with 𝑚𝑔. If we add the initial kinetic energy, the one-half 𝑚𝑣 initial squared, to both sides, we will now have an expression for the final kinetic energy. The final kinetic energy of the object is equal to 𝑚𝑔𝑑 plus one-half 𝑚𝑣 initial squared.

Before we plug in any values from our problem, we have to first make sure that they’re all in the correct units. Looking over our values, the only variable that needs to be converted is the mass, from grams to kilograms. One kilogram is equal to 1,000 grams. When we divide 400 grams by 1,000, we end up with a mass for our object of 0.400 kilograms. We replace the mass of the object with 0.400, 𝑔 with 9.8, 𝑑 with five, and the initial speed with four. When we multiply out our first term, we get 19.6. And when we multiply out our second term, we get 3.2. Adding these two terms together, we end up with a final kinetic energy of 22.8 joules, where joules is the standard unit for energy and is represented by the capital J. Using the work–energy principle, the body’s final kinetic energy just as it’s about to hit the ground is 22.8 joules.