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(II) The net force along the linear path of a particle of mass 480 g has been measured at 10.0 -cm intervals, starting at $x = 0.0 ,$ to be $26.0,28.5,28.8,29.6,32.8,40.1,46.6,42.2 ,$ $48.8,52.6,55.8,60.2,60.6,58.2,53.7,50.3,45.6,45.2,43.2$ $38.9,35.1,30.8,27.2,21.0,22.2 ,$ and $18.6 ,$ all in newtons. Determine the total work done on the particle over this entire range.

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$102 \mathrm{J}$

Physics 101 Mechanics

Chapter 7

Work and Energy

Work

Kinetic Energy

Potential Energy

Energy Conservation

Rutgers, The State University of New Jersey

University of Michigan - Ann Arbor

University of Washington

University of Sheffield

Lectures

04:05

In physics, a conservative force is a force that is path-independent, meaning that the total work done along any path in the field is the same. In other words, the work is independent of the path taken. The only force considered in classical physics to be conservative is gravitation.

03:47

In physics, the kinetic energy of an object is the energy which it possesses due to its motion. It is defined as the work needed to accelerate a body of a given mass from rest to its stated velocity. Having gained this energy during its acceleration, the body maintains this kinetic energy unless its speed changes. The same amount of work is done by the body in decelerating from its current speed to a state of rest. The kinetic energy of a rotating object is the sum of the kinetic energies of the object's parts.

04:31

(II) The net force exerted…

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(II) The force on a partic…

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02:34

02:59

The force acting on a part…

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Find the work done by the …

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04:50

07:33

02:30

So here, when you import that file, you Khun, import all the files and essentially used the formula in Excel and the formula is as follows. So the work applied. It's going to be equal to the sum Ah of I equals one. So it started one and then all the way to n minus one of 1/2 Times Force sub I plus force said I Plus one times Delta acts eso essentially approximating it as a trap is oId and we know that ex Delta X is going to be 10 centimeter increments or 0.1 acts and then the all the values are in Newton's already. So this will simply be the first term plus the next term times 1/2 times that increments. So essentially we're using The sum of trap is oi DS in order to find the work applied, and you could simply plug this into ah, you're Excel spreadsheet and plug in all of the forces. And then, if you were tio, apply this formula for all the cells. Ah, you would get the total amount of work done and again include that Delta X term that Delta X terms defines the increments so 10 centimeters or 100.1 meters. And when we apply this and we try to when we used the XL or any processing software to process the to calculate the work applied, we find that the work applied is going to be approximately 102 0.3 jewels. So essentially, you just need thio. Plug this into plug all the values into your Excel spreadsheet. Use this equation here. Remember to include the Delta X Term and then apply the equation to all the cells corresponding to the forces that are present again. All of them are already Nunes, so there's no need to convert. And the answer calculated should be approximately 102.3 jewels. That is the end of the solution. Thank you for watching.

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