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Just need a bit of help with this question and to make sure im solving it properly. Thanks

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A single conservative force acts on a 5.00 kg particle.

The equation F_x = (2x + 4) N describes this force, where x is in meters.

As the particle moves along the x axis from x = 1.40 m to x = 6.20 m, calculate the following:

(a) work done by the force

(b) change in the potential energy of the system

(c) kinetic energy of the particle at x=6.20m

if its speed is is 3.00 m/s at x =1.40m

Ek = 0.5mv^2

Ep = mgh

***A)***

Ok I started by integrating the force as ∫ 2x +4

= x^2 + 4x

I then evaluated the difference in area between the two bounds, namely

((6.2)^2 + 4*6.2 - (1.4)^2 + 4*1.4) which gave 55.68

Since this was the force I multiplied by the change in distance 6.2 - 1.4 = 4.8

So 55.68 * 4.8 = 267.264

***B)***

I figured that since the particle is always on the x-axis, the potential energy difference should be zero.

***C)***

At 1.4m, speed is 3m/s.

Ek = 0.5(5)(3)^2

= 22.5

They want Ek at 6.2m.

And now I can't decide what to do. Because I think I need the speed at 6.2m in order to find the energy at 6.2m, but I don't kno why the Ek at 1.4m was given. Or I might need to use work and total energy. Im really not too sure.

Any help to verify this would be appreciated.

Thanks

______________

A single conservative force acts on a 5.00 kg particle.

The equation F_x = (2x + 4) N describes this force, where x is in meters.

As the particle moves along the x axis from x = 1.40 m to x = 6.20 m, calculate the following:

(a) work done by the force

(b) change in the potential energy of the system

(c) kinetic energy of the particle at x=6.20m

if its speed is is 3.00 m/s at x =1.40m

Ek = 0.5mv^2

Ep = mgh

***A)***

Ok I started by integrating the force as ∫ 2x +4

= x^2 + 4x

I then evaluated the difference in area between the two bounds, namely

((6.2)^2 + 4*6.2 - (1.4)^2 + 4*1.4) which gave 55.68

Since this was the force I multiplied by the change in distance 6.2 - 1.4 = 4.8

So 55.68 * 4.8 = 267.264

***B)***

I figured that since the particle is always on the x-axis, the potential energy difference should be zero.

***C)***

At 1.4m, speed is 3m/s.

Ek = 0.5(5)(3)^2

= 22.5

They want Ek at 6.2m.

And now I can't decide what to do. Because I think I need the speed at 6.2m in order to find the energy at 6.2m, but I don't kno why the Ek at 1.4m was given. Or I might need to use work and total energy. Im really not too sure.

Any help to verify this would be appreciated.

Thanks

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