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A ball of mass five grams was moving in a straight line through a medium loaded with dust.
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The dust was accumulating on its surface at a rate of one gram per second.
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Find the magnitude of the force acting on the ball at time ๐ก equals five seconds, given that the displacement of the ball is expressed by the relation ๐ of ๐ก equals two-thirds ๐ก cubed plus ๐ก squared plus 7๐ก plus one ๐, where ๐ is a unit vector in the direction of the motion and the displacement is measured in centimeters.
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We want to solve for the magnitude of the force that acts on the ball at a particular time ๐ก equals five seconds.
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Weโll call that force magnitude capital ๐น.
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To solve for that force, it will help us to recall Newtonโs second law of motion.
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Newtonโs second law of motion is commonly written as force is equal to an objectโs mass times its acceleration ๐.
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This is a perfectly valid expression of the second law when the mass does not change with time and the speed of the motion is nonrelativistic.
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A more general expression for the second law is that the net force on an object is equal to the time rate of change of its mass times its velocity.
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In our scenario, we have a ball that starts out initially with a mass of five grams.
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Weโll call that value ๐ sub zero, yet it takes on mass as it gathers dust every second.
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In fact, if we were to write the mass of the ball as a function of time, it would be equal to its original mass five grams plus the time in seconds since each second that passes has one gram of mass to the ball.
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So in our scenario, the mass of our object does change in time.
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And therefore, we want to use this more general formulation of the second law.
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In writing this equation for our case, we see that ๐ as a function of ๐ก is something we know, but the velocity ๐ฃ is unknown currently.
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We can recall though that our velocity is equal to the time rate of change of position.
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And weโve been given a position vector for our ball.
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This means we can take the time rate of change of that position and substitute it in for ๐ฃ in our equation.
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When we take the time rate of change or derivative with respect to time of our position function, we get a result of two ๐ก squared plus two ๐ก plus seven in the ๐ hat direction.
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This is the expression for ๐ฃ that weโll put into our equation to solve for ๐น.
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But before we do, we can take the step of multiplying ๐ฃ by ๐ as a function of ๐ก, the mass of our ball, so that we can insert this expression entirely into our equation for ๐น.
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We discovered earlier that our mass as a function of time is equal to five plus ๐ก grams.
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When we multiply this term through with our velocity term, neglecting the units for the time being and grouping common powers of ๐ก, we find that this product is two ๐ก cubed plus 12๐ก squared plus 17๐ก plus 25.
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Thatโs ๐ as a function of ๐ก times the velocity of the particle ๐ฃ.
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So itโs this expression that weโll plug in to the parentheses to take the time derivative of to solve for ๐น.
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Weโre now set to take another time derivative.
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And when we do, we find the result of six ๐ก squared plus 24๐ก plus 17.
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This is close, but not quite the result that weโre looking for.
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We want to solve for the force ๐น magnitude at a particular ๐ก value โ that is when ๐ก is equal to five seconds.
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So weโll now plug in that particular value for ๐ก to solve for the force at that instant in time.
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With this value plugged in, when we calculate ๐น, we find its magnitude is 287.
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And the unit is dynes.
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This is the magnitude of the force that the ball experiences at time ๐ก equals five seconds.