00:03This time we will solve problem 13.69 from the book, Dynamics, by the author Beer.
00:11Regarding the topic of work and energy, the statement reads as follows.
00:15A spring is used to stop a 50-kilogram package moving down a slope
00:21of 20 degrees.
00:22The spring has a spring constant K, equal to 30 kilonewtons per meter, and is supported by cables, so that
00:30Initially it is compressed 50 millimeters.
00:33If the speed of the package is known to be 2 meters per second when it is 8 meters away
00:38of the spring,
00:40and if the coefficient of kinetic friction between the package and the inclined plane is 0.2,
00:46Determine the maximum additional deformation of the spring, to bring the package to rest.
00:53Returning to the problem, we first made a diagram showing the reaction generated by the floor and the block.
01:00and the force of weight W.
01:04We apply the sum of forces on the axis where the reaction acts.
01:08The sign convention is proposed as positive in an ascending manner; two forces act on this axis.
01:15The normal and the projection of the weight W, which, being the adjacent leg, uses the trigonometric function cosine.
01:24We solve for the normal force, and from theory we know that the frictional force is equal to the product of the
01:30normal by the coefficient of friction.
01:33We substitute the expression for the normal force, obtaining an equation that relates us to the frictional force,
01:39with the coefficient of friction, and the weight, which was replaced by the product of the mass and the
01:44gravity.
01:47To solve the problem, we will use the concept of work and energy.
01:52Given an initial velocity, the kinetic energy of point 1 is equal to one half of the mass times
01:57the speed squared.
02:00The problem asks us to perform a calculation, in which the block stops, meaning that it does not carry out
02:06speed.
02:07Therefore, the kinetic energy of point 2 is equal to zero.
02:12During the block's displacement, there are three forces that generate work.
02:16Weight, friction force, and spring force; the sum of these three terms gives us the
02:23total work done.
02:25The work done by weight is equal to mass times gravity times height.
02:31where the sign is positive, because the weight vector has the same direction as the block's displacement.
02:37To calculate the height, we draw a diagram highlighting a right triangle in the figure provided by the problem.
02:47The height the block moves ends up being the opposite side with respect to the angle,
02:52For this reason we use the trigonometric sine function,
02:56to relate height to displacement on the inclined plane,
03:01We substitute this equation into the work of weight,
03:04obtaining the equation shown on the screen.
03:09The work done by friction is equal to the frictional force,
03:14which multiplies the distance the block travels on the inclined plane.
03:18The minus sign,
03:20This is because the direction of the force is opposite to the displacement of the block.
03:26from the analysis carried out at the beginning of the problem,
03:28We know that in the force of friction,
03:30is equal to the term that is observed,
03:36we substitute in the work of friction,
03:38obtaining the following,
03:41And finally, from theory we know that the force of a spring,
03:45is equal to the equation shown on the screen,
03:50We substitute each of the expressions obtained in work 1 to 2,
03:56the term mass times gravity multiplied by 8 plus delta of x,
04:00we can factor it,
04:03We also know that the difference in x is equal to the deformation 2 minus the deformation 1.
04:11From the equation obtained, we know the magnitude of some terms,
04:15We verify the consistency of units,
04:21We replace and operate,
04:24Now, we substitute the kinetic energy from points 1 and 2,
04:28and the total work obtained,
04:30in the work-energy equation,
04:35The problem statement gives us the mass and the speed as given data.
04:40We verify the consistency of units,
04:43we replace,
04:44we operate,
04:45we combine like terms,
04:48resulting in the general form of a quadratic equation,
04:54we found the two roots,
04:55and the value that will be used will be the positive one,
05:01The problem asks us to calculate the maximum additional deformation,
05:05This value is equal to the difference in deformations,
05:09where both values are already known,
05:13we replace,
05:14we operate,
05:15and Delta X,
05:16is equal to 174.39 millimeters,
05:20being the solution to the problem,
05:23Well, if it's helpful to you,
05:25Don't forget to subscribe!
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