MAT 224 Elementary Linear Algebra and Differential Equations

MAT 224: Elementary Linear Algebra and Differential Equations: One end of a rubber band is fixed at a point A. An object of 1-kilogram mass, attached to the other end, stretches the rubber band vertically to point B in such a way that the length of AB is 16 centimeters greater than the natural length of the band. If the mass is further drawn to a position 8 centimeters below B and released, what its velocity (if we neglect resistance) as it passes point B?

Solution

Data.

ab = 16, mass = 1 kg

ab\' = 24, mass = 1.5 kg

resistance = 0

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Acceleration is the derivative of velocity with respect to time;

a(t) = v*t

so;

v (t) = a*t

v´(t) = a

In this case Newton\'s Second Law of Motion applies in the following way: It says to us that \"the mass of an object times its acceleration is equal to the sum of the forces acting on it\". We got two forces, gravity and resistance. The force of gravity is the weight of the object: 9.8*mass. Air resistance can be handled in several ways, we choose to neglect it here!

Recalling,

mass * a = gravity + resistance

then;

mass*v\'(t) = -9.8*mass + 0

mv\' = -9.8m

v´(t) = -9.8

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Therefore velocity is 9.8 m/s as it passes point b after further drawn.