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A body of mass 96 kilograms was moving in a straight line at 17 meters per second.
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A force started acting on it in the opposite direction to its motion.
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As a result, over the next 96 meters, its speed decreased to 11 meters per second.
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Using the work–energy principle, determine the magnitude of the force.
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In the problem, we are instructed to use the work–energy principle.
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In equation form, this states that the net work down on an object is equal to the change in kinetic energy of the object.
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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.
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Change in kinetic energy is the final kinetic energy minus the initial kinetic energy.
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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.
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We can replace kinetic energy final with one-half 𝑚𝑣 final squared and kinetic energy initial with one-half 𝑚𝑣 initial squared.
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To isolate the force, we can divide both sides of the equation by 𝑑, which will cancel out the displacement on the left side.
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Now that we have an expression for our force, we can substitute in the values from our problem.
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We use 96 for the mass, 11 for the final velocity, 17 for the initial velocity, and 96 for the displacement.
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When we multiply out our first term, one-half times 96 times 11 squared, we get 5,808.
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Multiplying out the second term of one-half times 96 times 17 squared, we get 13,872.
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When we subtract our numerator and divide by our denominator, we get a force of negative 84 newtons.
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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.
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Using the work–energy principle, the magnitude of the force acting on the object is 84 newtons.