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A car (S driving on level highway It /s acted upon by Ihe (ollowing forces downwrard gravitational force of 12 kN, an upward contact force due t0 Ihe roud of 12 kN,...

Question

A car (S driving on level highway It /s acted upon by Ihe (ollowing forces downwrard gravitational force of 12 kN, an upward contact force due t0 Ihe roud of 12 kN, anolhier conlact force due t0 the rOad of - KN dlrected West; and drag force due to alr resistance ol 5 kN directed East What Is the net force acting On the car?

A car (S driving on level highway It /s acted upon by Ihe (ollowing forces downwrard gravitational force of 12 kN, an upward contact force due t0 Ihe roud of 12 kN, anolhier conlact force due t0 the rOad of - KN dlrected West; and drag force due to alr resistance ol 5 kN directed East What Is the net force acting On the car?



Answers

A truck driving on a level highway is acted on by the following forces: a downward gravitational force of $52 \mathrm{kN}$ (kilonewtons); an upward contact force due to the road of $52 \mathrm{kN} ;$ another contact force due to the road of $7 \mathrm{kN},$ directed east; and a drag force due to air resistance of $5 \mathrm{kN}$, directed west. What is the net force acting on the truck?

So here we have a truck and we could say that the sum of forces in the Z direction here we have the gravitational force from the truck and rather, we could say the normal force from the, uh pavement would actually be positive minus the gravitational force of the truck. And of course, there isn't any acceleration in the Z direction, so that's gonna be equal to zero. The truck is not floating, of course. And so there would be zero Newton's and so the the other forces. Now we have eps of one equaling seven killer Newtons on this would be east and we have said to this would be equal to five killing Newtons and this would be west. So we can then say that the net force in the X direction the sum of forces in the extraction this would be equal to seven killing. Newton's were considering east to be positive minus five, killing Newtons giving us to killing Newtons and this would be east, considering it's positive and the sum of forces in the Y direction, of course, is again zero Newtons. So we can then say that the net force acting on the truck is going to be equal to to kill and Newton's. And we can say I had to the positive X direction or again East. That is the end of the solution. Thank you for

So let's start by making a free body diagram. So we've got our car and there's the weight off the car. Since the masses 1200 kilograms of this is the wit times it by nine point it gives me the wheat is 117 60 Newton's, then untold that there's a normal force. And then there's a force of friction. So that's the F. And that's equal to 524 Newtons. And then there's a forward force F, which I do not know, And then the question say, is that the network done by all the forces is 150 Killer Jules. Now we know that if the angle of tea incline off road is five degrees and this angle will also be five degrees, which means that the weight actually makes an angle of 95 degrees with the displacement and the displacement is up the hill, and that's 290 meters. So we are told. So let's first of all, say that the network done in this case is equal to the work done by the weight plus the work and by the normal force, plus the work and by friction, plus the worked on by the Applied Force. Now we know that worked, and by normal force would be zero, because the normal force is perpendicular to the direction of motion. And the question to set tells me that the network done is 150 killer jewels. So the right the left side of the equation is 150 times 1000. And now let's find the work done by the force off crowd. You're working by the weight, so that's going to be 11760 times 290 times co sign off 95 degrees. So work is force times, displacement times the angle between them. Ah, force of friction is in the opposite direction. Off the motions or worked on. The friction would be negative 5 24 times to 90. So work is equal to force time displacement, plus the work done by the force. We do not know the magnitude of the force, but we do know that the force is applied over a distance of 290 meters. So let's go ahead and solve this equation for F and solving for F gives me 1.55 times 10 to the power three news

In both cases, acceleration equals zero. So the man force equals mass times. Acceleration equals zero. So in both guesses, that is known that force acting on the

In both cases, acceleration equals zero. So the man force equals mass times. Acceleration equals zero. So in both guesses, that is known that force acting on the


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