Video: Finding the Magnitude of the Force Acting on a Moving Body at a Given Time Where Dust Is Accumulating on the Body While Moving given Its Displacement Expression

Video Transcript

A ball of mass five grams was moving in a straight line through a medium loaded with dust. The dust was accumulating on its surface at a rate of one gram per second. Find the magnitude of the force acting on the ball at time 𝑡 equals five seconds, given that the displacement of the ball is expressed by the relation 𝑠 of 𝑡 equals two-thirds 𝑡 cubed plus 𝑡 squared plus 7𝑡 plus one 𝑐, where 𝑐 is a unit vector in the direction of the motion and the displacement is measured in centimeters.

We want to solve for the magnitude of the force that acts on the ball at a particular time 𝑡 equals five seconds. We’ll call that force magnitude capital 𝐹. To solve for that force, it will help us to recall Newton’s second law of motion. Newton’s second law of motion is commonly written as force is equal to an object’s mass times its acceleration 𝑎. 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. 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.

In our scenario, we have a ball that starts out initially with a mass of five grams. We’ll call that value 𝑚 sub zero, yet it takes on mass as it gathers dust every second. 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.

So in our scenario, the mass of our object does change in time. And therefore, we want to use this more general formulation of the second law. In writing this equation for our case, we see that 𝑚 as a function of 𝑡 is something we know, but the velocity 𝑣 is unknown currently.

We can recall though that our velocity is equal to the time rate of change of position. And we’ve been given a position vector for our ball. This means we can take the time rate of change of that position and substitute it in for 𝑣 in our equation. When we take the time rate of change or derivative with respect to time of our position function, we get a result of two 𝑡 squared plus two 𝑡 plus seven in the 𝑐 hat direction. This is the expression for 𝑣 that we’ll put into our equation to solve for 𝐹.

But before we do, we can take the step of multiplying 𝑣 by 𝑚 as a function of 𝑡, the mass of our ball, so that we can insert this expression entirely into our equation for 𝐹. We discovered earlier that our mass as a function of time is equal to five plus 𝑡 grams.

When we multiply this term through with our velocity term, neglecting the units for the time being and grouping common powers of 𝑡, we find that this product is two 𝑡 cubed plus 12𝑡 squared plus 17𝑡 plus 25. That’s 𝑚 as a function of 𝑡 times the velocity of the particle 𝑣. So it’s this expression that we’ll plug in to the parentheses to take the time derivative of to solve for 𝐹.

We’re now set to take another time derivative. And when we do, we find the result of six 𝑡 squared plus 24𝑡 plus 17. This is close, but not quite the result that we’re looking for.

We want to solve for the force 𝐹 magnitude at a particular 𝑡 value — that is when 𝑡 is equal to five seconds. So we’ll now plug in that particular value for 𝑡 to solve for the force at that instant in time.

With this value plugged in, when we calculate 𝐹, we find its magnitude is 287. And the unit is dynes. This is the magnitude of the force that the ball experiences at time 𝑡 equals five seconds.