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A $1200-\mathrm{kg}$ car is being driven up a $5.0^{\circ}$ hill. The frictional forceis directed opposite to the motion of the car and has a magnitude of $f=524 \mathrm{N}$ . A force $\overrightarrow{\mathbf{F}}$ is applied to the car by the road and propels the car forward. In addition to these two forces, two other forces act on the car: its weight $\overrightarrow{\mathbf{W}}$ and the normal force $\overrightarrow{\mathbf{F}}_{\mathrm{N}}$ directed perpendicular to the road surface. The length of the road up the hill is 290 $\mathrm{m}$ . What should be the magnitude of $\overrightarrow{\mathbf{F}},$ so that the net work done by all the forces acting on the car is $+150 \mathrm{kJ}$ ?

$F=1.55 \times 10^{3} \mathrm{~N}$

Physics 101 Mechanics

Chapter 6

Work and Energy

Work

Kinetic Energy

Cornell University

University of Michigan - Ann Arbor

Hope College

Lectures

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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

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