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in this video we're gonna focus on some.physics problems that will help us to.calculate the net force in the X.direction or in the horizontal direction.so let's start with this problem a 5.kilogram box is pulled to the right by a.horizontal force of 200 Newtons a.constant frictional force of 60 Newtons.opposes it Part A calculate the net.horizontal force acting on the box so.let's draw a picture so let's say this.is the 5 kilogram box and it's being.pulled to the right by a 200 Newton.force now there is a constant frictional.force that opposes that friction always.opposes motion so if the box is moving.to the right friction is going to be.directed towards the left and the.frictional force is 60 Newtons so let's.write an expression for the net force so.then that force in the x-direction is.going to be equal to F because this is.directed in the positive x-direction.this F is positive now this one is.directed in the negative x-direction so.it's going to be negative F so then that.horizontal force in the X Direction is.the difference between these two forces.and so that's gonna be 200 minus 60 so.the net horizontal force is simply 140.Newtons so this is the answer to Part A.now what about Part B how can we now.calculate the acceleration acted on his.box now Newton's second law states that.the net force be in the X direction or.in the Y Direction is equal to MA.mass times acceleration and it also.states that if the net force is in the X.direction the acceleration has to be in.X direction if the net force is in the Y.direction the acceleration has to be.the y direction the acceleration and.force factors will always be in the same.direction now the net force we know is.140 the mass of the block is 5 kilograms.so it's a funny acceleration in a.horizontal direction it's going to be.140 divided by 5 and so and that comes.out to be 28 meters per second squared.so that's the answer for Part B so Part.C how far will the box travel after 15.seconds so that's simply a key to Matic.problem so let me just get rid of this.stuff.now let's write down the important.information that we have so we have the.acceleration it's 28 meters per second.squared and we know the time the time is.15 seconds now we don't have an initial.speed so we're going to assume that the.box accelerated from rest so the initial.speed we're going to assign a value of 0.to it our our goal is to find the.distance traveled by the box.what kinematic form ladino has these 4.variables here's the equation that you.need the displacement which is going to.be the same as distance in this problem.because it's moving in one direction.it's not changing direction the.displacement is gonna be V initial T.plus 1/2 a t-square the initial speed is.zero so it's gonna be one-half times 28.times 15 squared half of 28 is 14 15.squared is 225 so let's multiply 14 by.225 and so the displacement is 3000 150.meters which is three point one five.kilometers so that's how far the box is.going to travel if these two forces are.applied to it.now let's work on another problem a 12.kilogram box is pulled to the right by a.350 Newton force that is 30 degrees.above the horizontal and a constant.frictional force of 120 Newton's opposes.it so let's draw a picture it's going to.be similar to the last picture but.slightly different as well so it's not.exactly the same so here's the.frictional force that opposes it and.it's a hundred and twenty Newtons now we.got another force that is 30 degrees.above the horizontal and that force.it's 350 Newtons so with this picture.what is the net horizontal force acting.on the box.feel free to pause the video to work on.this example now because the applied.force is at an angle it's going to have.an X component and a Y component our.goal is to find the net horizontal force.so we need only concern ourselves with.the X component of this force so the net.force in the X Direction is going to be.f of X which is in the positive x.direction and this one is in a negative.x direction so it's gonna be minus.lowercase F now what is FX FY is F sine.theta and F X is f cosine theta so this.is what we need in this example so this.is going to be F times cosine theta.minus F so that's gonna be 350 times the.cosine of 30 degrees minus 120 now make.sure your calculator is in degree mode.so you can get the right answer.350 times cosine 30 that's about 300 and.3.1 and if we subtract that by 120.that will give us a net force in the.x-direction of a hundred and eighty 3.1.newtons so that's the answer to Part A.now let's move on to Part B so let's get.rid of most of the stuff that we have.here to find the acceleration we could.use this equation just like we did.before then that force is going to be M.times a so then that force is one.hundred eighty three point one the mass.of the box is 12 kilograms and so the.acceleration is 183 point one divided by.12 and so the acceleration in the.x-direction is 15 point 26 meters per.second squared.so now we can focus on Part C now that.we have the answer it's a Part A so what.is the final speed of the box after it.traveled a distance of 200 meters so.let's make a list of what we have we.have the acceleration we're looking for.the final speed we're going to assume.that the initial speed is zero and the.displacement is 200 meters so the.equation that has these four variables.is this one V final squared is equal to.V initial squared plus two ad so Venus.show is zero the acceleration is fifteen.point 26 and D is 200 so 2 times 15.point 26 times 200 that's 6,000 104 so.that's equal to the square of the final.speed so now we need to take the square.root of both sides to get the final.speed so the square root of 60 104 is 78.point one meters per second so that's.the final speed of the box.after it traveled a distance of 200.meters so this is gonna be the last.problem of this video a 1200 kilogram.car speeds up from 25 meters per second.to 60 minutes per second in five seconds.what is the acceleration of the car well.let's make a list of what we have we.have the initial speed of 25 we have a.final speed of 60 and a time value of 5.what is the acceleration what equation.has these four variables perhaps you've.seen this one V final is equal to V.initial plus 18 V final is 60 V initial.is 25 we're looking for a and T is 5 so.first we got a subtract both sides by 25.60 minus 25 is 35 so 35 is equal to 5 a.so if we divide both sides by 5 35.divided by 5 is 7 so the acceleration is.7 meters per second squared so now that.we have the acceleration we can use that.to calculate the net force so Part B.what is the net force acting on the car.then that force in a horizontal.direction is the mass times the.horizontal acceleration and makes sense.that the car will be traveling in.horizontal direction so F equals MA it's.just gonna be 1200 times an acceleration.of 7 now 12 times 7 is 84 so this is.going to be 8400 Newton's so that's the.net force in the X direction so that's.the answer to Part B now Part C a real.part B twice but this is supposed to be.Part C if the car experience is a.constant frictional force of 3,500.Newtons what is the average force.exerted by the engines on the car so.let's draw a little picture.so here's the car as a mass M and the.engines is applying a force that's going.to drive the car in the forward.direction and friction is going to.oppose it so just like before the net.force is going to be the difference.between the applied force and the.frictional force which I like to use.lowercase F to represent that our goal.is to find the force exerted by the.engines on the car so that's F so if.we're trying to solve for F we'll need.to add friction to both sides of the.equation so the applied force I'm going.to put an AM excellent the force applied.by the engines is going to be the net.force plus the frictional force the net.force is 8400 and the frictional force.is 3,500 which adds up to eleven.thousand nine hundred Newton's so let's.make sense of this so the box represents.the car the engines applies a force of.eleven thousand nine hundred Newton's to.drive the car to the right friction.slows it down by thirty five hundred.Newtons so the net force in the.x-direction is the difference between.these two which is 8400 minutes so.hopefully you see how these three.numbers work out but this is the answer.to the question that we're looking for.which is the applied force the force.exerted by the engines on the car.

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What is the right way to fill out Two-Earners Worksheet tax form?

You might have found answer, but I am posting a link to help others who hit this question: http://www.irs.gov/Individuals/IRS-Withholding-Calculator This gives a fairly accurate guidance

How can I fill out Google's intern host matching form to optimize my chances of receiving a match?

I was selected for a summer internship 2016. I tried to be very open while filling the preference form: I choose many products as my favorite products and I said I'm open about the team I want to join. I even was very open in the location and start date to get host matching interviews (I negotiated the start date in the interview until both me and my host were happy.) You could ask your recruiter to review your form (there are very cool and could help you a lot since they have a bigger experience). Do a search on the potential team. Before the interviews, try to find smart question that you are Continue Reading

What is net resultant force?

The resultant force is the vector sum of an arbitrary number of force vectors. The net force is the resultant force of ALL the force vectors acting on a system. In the case cited above the net force is given by: In the special case where we then have which reduces the net force to:

What is the formula of resultant force?

​ ​ Formula R = (p^2 +q^2 + 2×p×q×cos(θ))^(0.5) Where R = Magnitude of resultant vector α = Direction of resultant vector P = Magnitude of vector P Q = Magnitude of vector Q θ = Angle between two vectors https://www.easycalculation.com/physics/classical-physics/learn-resultant-vector.php

What is the net force of an object?

Read the following and see if this helps Newton's first law states that every object will remain at rest or in uniform motion in a straight line unless compelled to change its state by the action of an external force.

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