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You apply a constant force $\overrightarrow{F} =(-68.0 \, \mathrm{N})\hat{\imath} +(36.0 \, \mathrm{N})\hat{\jmath}$ to a 380-kg car as the car travels 48.0 m in a direction that is 240.0$^\circ$ counterclockwise from the +$x$-axis. How much work does the force you apply do on the car?

$$+135 \mathrm{J}$$

Physics 101 Mechanics

Chapter 6

Work and Kinetic Energy

Section 1

Work

Kinetic Energy

Cornell University

Rutgers, The State University of New Jersey

University of Washington

University of Winnipeg

Lectures

03:47

In physics, the kinetic en…

02:08

In physics, work is the tr…

00:51

A small car of mass 380 kg…

01:29

A small car (mass 380 $\ma…

00:35

(a) A car with a mass of $…

01:22

The velocity of a $625-\ma…

02:36

A $1200-\mathrm{kg}$ car i…

01:04

An $1850 \mathrm{kg}$ car …

0:00

A 970.-kg car starts from …

03:55

Two forces are applied to …

00:33

04:15

A force $\overrightarrow{F…

06:37

A car weighing 12,500 N st…

00:41

What is the acceleration o…

02:12

When you take your 1900 -k…

01:23

ssm A $1380-\mathrm{kg}$ c…

15:31

Suppose the 2.0-kg model c…

02:13

Work A car drives 500 $\ma…

01:33

A mechanic pushes a $2.50 …

02:33

6.30. Suppose the $2.0-\ma…

04:51

The wheels of a midsize ca…

02:17

A curve in a road forms pa…

hi once again welcome to a new problem. This time was still dealing with forces and energy. There's always a relationship involving forces and energy. Remember, walk down. What down is a form of energy, which is force times displacement. So we have a case where we have a car that's traveling in a specific direction precisely two hundred and forty degrees counterclockwise from the X direction. That means it's traveling in the southwest. This is not so east and west, so there is an angle or sixty degrees right there between western south. That's not a big deal. For now, we know that thie call itself happens to have a mass off three hundred and eighty kilograms on DH. The distance it travels in this direction is forty eight meters in that direction. So first direction at that angle. Well, given the force in terms of vectors, the force vector is negative. Sixty eight point zero Newtons I had plus, that is six point zero Newton's jihad. That's the force that's pushing the car in the Southwest direction two hundred and forty degrees from the X direction. So that was the displacement eyes given in terms ofthe iron J the displacement is just given in terms of meters. So for the eight meters. But we can change that in terms ofthe NJ by saying this is forty eight point zero meters times co sign off two hundred and forty point zero degrees. I heard s so we're doing a transformation where we change the displacement also in terms ofthe actors. Because remember the forces in terms ofthe actors also s o, we want to maintain a consistent convention. So sign off. Two hundred and forty point zero degrees Jihad. That gives us a displacement off. Negative. Twenty eight point five meters. I heard. Ah, plus rather minus. Not a plus. Minus miners. Um, this is this is now going to be minus jihad of forty nine point three six meters jihad. So in the next page, we're going to compute the walk down by the force in pushing the car. Oh, in moving the car in the direction in the southwest direction two hundred forty degrees counterclockwise from the x axis from the positive ex axes. So then what? Bom dun by force, Uh, w happens to be f with the symbol for vector on top of it on a CE for the displacement factor. So the force is negative. Sixty eight point zero Newton's, I heard. Plus, uh, thirty six one zero Newtons. Jihad times negative. That's the displacement in terms of factors we got in the previous page, as you can recall, uh, I had of minors forty nine point three six meters. J. HUD. So we're multiplying those, too. Uh, negative. Sixty eight point zero Newtons Times negative twenty eight point five. Um, we need to go back a little bit. Some Something happened in terms of our numbers in terms of our numbers. Call sign. We want to cross check if we got this. This number's right. So, you know, we have forty eight times co sign off two hundred and forty, and that gave us negative twenty four. So one of these super precise negative twenty four. Ah, that's the displacement in the eye direction. So that happens to be negative. Twenty four on point zero meters. I had, and then the other one wass forty eight times sign off two hundred and forty. And that gives us negative forty one points. So this would be a minus. This is just going to be a minus. Negative for the one point five seven meters J hut. Okay, so when we go back to the other, Paige, we need to change these numbers to conform with the results. We had initially, you know, the forces were negative. Sixty eight mutants and thirty six. So this one, this values right here. We're going to change. You have to change them to make sure they line up without results on the previous page. So this one happened to be just to make sure. Negative twenty four meters to the eye. Hard minus forty one point five seven meters. Jae ha. So this is twenty eight. This is gonna change to twenty four meters. We have to change that. This's twenty four meters, Um, and then plus thirty six point zero Newtons times negative forty one point five seven leaders final answer becomes We have to put those two together sixty eight times. Twenty four. That gives us sixteen thirty two jewels. Remember DS positive? Because this one is negative and that's negative. And then thirty six times. Forty one one five seven. That's negative. Fourteen and six point five two. When we subtract those two together, we get one hundred, one hundred and thirty five point for eight. Jules. So, once again, thank you very much. I hope you enjoyed the video. If you have any comments filled free too, respond or cents under my way. Way had a force pushing a car in the direction of the ex direction of two hundred positive ex direction of two hundred and forty our degrees at a distance off create. We had to change the displacement into a vector by multiplying by forty eight and the unit factors. And so that gives us a negative twenty four meters in the air. Negative. Forty one point five seven. Video is, uh, in the J. Huh? Product force and displacement that give us the work done by the Force. A net force of one hundred A net energy off hundred and thirty five point four eight. Jules. Okay, Thanks. Bye.

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