A body of mass 96 kilograms was moving in a straight line at 17 meters per second. A force started acting on it in the opposite direction to its motion. As a result, over the next 96 meters, its speed decreased to 11 meters per second. Using the work–energy principle, determine the magnitude of the force.
In the problem, we are instructed to use the work–energy principle. In equation form, this states that the net work down on an object is equal to the change in kinetic energy of the object. Recalling that work is defined as force times displacement, where the force is parallel to the displacement, we can expand out our formula, replacing 𝑊 with 𝐹 times 𝑑. Change in kinetic energy is the final kinetic energy minus the initial kinetic energy. We should remember that the kinetic energy of an object is equal to one-half 𝑚𝑣 squared, where 𝑚 is the mass of the object and 𝑣 is the speed of the object. We can replace kinetic energy final with one-half 𝑚𝑣 final squared and kinetic energy initial with one-half 𝑚𝑣 initial squared. To isolate the force, we can divide both sides of the equation by 𝑑, which will cancel out the displacement on the left side.
Now that we have an expression for our force, we can substitute in the values from our problem. We use 96 for the mass, 11 for the final velocity, 17 for the initial velocity, and 96 for the displacement. When we multiply out our first term, one-half times 96 times 11 squared, we get 5,808. Multiplying out the second term of one-half times 96 times 17 squared, we get 13,872. When we subtract our numerator and divide by our denominator, we get a force of negative 84 newtons. We are asked to find the magnitude of the force, and therefore we do not need the negative sign as that tells us the direction. Using the work–energy principle, the magnitude of the force acting on the object is 84 newtons.