In four previous articles, I described how I teach motion with constant acceleration and Newton's second law. In these articles, I explained how I try to get students to solve the problems of Physics in terms of fundamental principles. You can find the details of my arguments in the foregoing articles.
Due to the limitations of Equation Editor, I use some unusual rating. This notation is also described in the preceding articles. For the first time here, I represent an angle with a variable. I represent "Theta" by th and "phi" by phi.
In this article, I describe how to use the theorem of energy work to solve the problems of physics. As usual, I will explain my approach with examples. Please note how each important step in my problem solutions begins with the statement of principles. It must be preceded by determine formulas work for various forces. These formulas can find in any physical book introduction.
Problem. A case of mass m is driven by a force p applied at an angle below the horizontal th. The coefficient of friction of sliding between the caisse and the floor is the United Kingdom. If the Fund starts at rest, what is its speed after he pushed a distance D?
Analysis. We use a framework of inertial reference with the horizontal axis and axis is vertical x. The initial point is represented by i and the end point by u. The caisse is touch: (i) that it grows (force P); (ii) the floor (normal force n and friction force UkN). (iii) the weight of the caisse is MG. With the help of a free body diagram, we can apply Newton's second law in the vertical direction.
Newton's second law
Sum (FY) = MAy
N MG - Psin (th) = 0
So, N = MG + Psin (th).
Now, we can apply the work-energy theorem.
Work-energy theorem
Work (P), Work (N) + Work (UkN) Work (MG) = M(Vu**2)/2-M(Vi**2)/2.
PDcos (th) + 0 - UkND + 0 = (MVu * 2) / 2-(IVM * 2) / 2.
Finally, with N = MG + Psin (th) and Vi = 0, we
PDcos (th) - Uk (MG + Psin (th)) D = M(Vu**2)/2.
Thus, Vu * 2 = 2 (Pcos (th) - Uk (MG + Psin (th))) D) / M.
Problem. Massless with constant force spring K = 100 N/m hangs from the ceiling, as shown in the sketch. A ball of mass M = 1,0 kg is attached at the end of spring and released. What is the speed of the ball to the point where he fell a distance D = 5.0 cm?
Analysis. We use a vertical axis x from the free end of the spring without stretching and positive up to study the movement of the ball. The only forces on the ball are KX - due to spring and MG due to gravity. The ball is released when the spring is not stretched, its initial position and velocity are Xi = 0 and Vi = 0. We want that velocity of the ball saw when he fell to 5.0 cm distance, where its position is Xu =-5.0 cm. With the work-energy theorem and the equations for the work, we have
Work-energy theorem
W (Spring) + W (gravity) = M(Vu**2)/2-M(Vi**2)/2.
K(Xu**2)/2-K(Xi**2)/2 + MG (Xu - Xi) = M(Vu**2)/2-M(Vi**2)/2.
You can do arithmetic you use simply K = 100 N/m, Xi = 0, Xu = 0.050 m, Vi = 0 and M = 1,0 kg. Willl you find light = 0.85 m/s.
Here we have two solutions more Physics problems that follow the same basic introduced (and highlighted) method in one-dimensional kinematics, object physics introduction first. For these issues resolved, calculations start with a statement of principle (Newton's second law and premier work-energy theorem and the principle of labour-power in the second). This is my basic message to students: start all solutions of the problem with a declaration of principles applicable in physics.
Dr. William Moebs is physical retired professor who has taught at both universities: Indiana - Purdue Fort Wayne and Loyola Marymount University. You can see hundreds of examples illustrating how stressed the fundamental principles by visiting physical support.
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