A 1.50-kg book is sliding along a rough horizontal surface. At po...

Question

A 1.50-kg book is sliding along a rough horizontal surface. At point A it is moving at 3.21 m/s, and at point B it has slowed to 1.25 m/s. (a) How much work was done on the book between A and B?

Accepted Answer

Hey everyone today, we're dealing with the problem that requires the work energy theorem. So, in our questions then we're being told to calculate the work needed to be put into a box that is traveling between two points X and Y along an irregular surface. So an irregular surface, an irregular horizontal surface is a horizontal surface that may have some ridges and dips. It is not perfectly horizontal, not perfectly flat throughout, but you can still push an object along it. Now we're traveling between two points X&Y. And we also know a few things. We have a box That weighs two kg at point X. It has a velocity velocity at X is equal to Uh three m/s, And the velocity at Point Y is equal to 2.25 m/s. And the direction of travel is to the right, it's going from point X to point Y. So using mass and velocity, we can use the work Energy theorem because recall the work energy theorem states that the network in the system is equal to the final kinetic energy minus the initial kinetic energy, which in this case would be the kinetic energy at Y. Kinetic energy at point y minus the kinetic energy at point X. Again, we can also recall that kinetic energy is equal to 1/ M. V squared. So, substituting this into our work energy theorem, we get that is the kinetic energy at point Y, which will use the velocity in y minus the kinetic energy at point X, which uses the velocity of point X squared. So now that we have this, we can actually factor out and simplify this a little bit. We can factor out the one half M term. And we get that It's 1/2. Multiplied by V Y squared minus v X squared. Substituting in our values here. Now We get that this is equal to 1/2. The box is two kg. It is too kilograms. The y squared is simply 2.25 m per second squared, which I'll just simplify the 2.25 squared for the sake of simplicity minus the initial velocity, which is 3.0 m per second squared and simplifying all of this, we get a final answer Of -3.94 jewels. Therefore the work required to put into the box as it travels between points X and Y is answer choice B -3.94 Jules. I hope this helps. And I look forward to seeing you all in the next one.

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