A particle moving along the x-axis is acted upon by the force whose functional form appears below. The velocity of the particle at x=0x=0 is v=6.0~ ext{m/s}v=6.0 m/s. Find the particle’s speed at x=(a)2.0m (b) 4.0m , (c)10.0 m, (d) Does the partciale turn around at some point and head back towards the origin ? (e) Repeat part (d) if v=2.0 m/s at x=0

Physics, 24.02.2020 16:34, jholland03

A particle moving along the x-axis is acted upon by the force whose functional form appears below. The velocity of the particle at x=0x=0 is v=6.0~\text{m/s}v=6.0 m/s. Find the particle’s speed at x=(a)2.0m (b) 4.0m , (c)10.0 m, (d) Does the partciale turn around at some point and head back towards the origin ? (e) Repeat part (d) if v=2.0 m/s at x=0

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answered: oshagie

The given force distance graph of the particle gives the work done by the

force on the particle.

(a) The particle's speed at x = 2 is approximately 5.568 m/s.(b) Velocity of the particle at x = 4 is approximately 5.22 m/s.(c) Velocity of the particle at x = 10 is approximately 6.403 m/s.(d) The particle does not turn around or head back towards the origin(e) Yes the particle turns around.

Reasons:

(a) From Newton's second Law of motion, we have;

$Acceleration\ a = \dfrac{Force}{Mass}$

From the given graph, we have;

Force acting on the particle from x = 0 to x = 2 m is -5 N

Therefore;

$\dfrac{-5 \ N}{4 \ kg} = -1.25 \ m/s^2$

The velocity at x = 2, is v² = u² + 2·a·s

s = Distance = 2 m

u = Initial velocity = 6 m/s

∴ v² = 6² - 2 × 1.25 × 2 = 31

v = √(31) ≈ 5.568

The particle's speed at x = 2, v ≈ 5.568 m/s

(b) The work done by the force from x = 2 to x = 4 is given by the area of

the trapezium that has a base from x = 2 to x = 4, as follows;

$Work \ done \ on \ the \ particle, W = \mathbf{\dfrac{(2 \, m + 1 \, m) \times (-5 \, N)}{2}} = -7.5 \, J$

Kinetic energy of the particle at x = 4 is therefore;

K.E. = 0.5 × 4 × v² = 0.5×4×31 - 7.5 = 54.5

The velocity at x = 4, v, is therefore;

$v = \sqrt{\dfrac{54.5}{0.5 \times 4} } = \sqrt{27.25} \approx 5.22$

The velocity of the particle at x = 4, v ≈ 5.22 m/s

Bases of trapezium, a = 10 m - 4 m = 6 m

b = 10 m - 5 m = 5 m

$Area \ of \ trapezium = Work \ on \ particle, W = \dfrac{(6 \, m + 5 \, m) \times (5 \, N)}{2} = \mathbf{27.5 \, J}$

Energy in particle, K.E. = 0.5 × 4 × v² = 0.5×4×27.25 + 27.5 = 82

$v = \sqrt{\dfrac{82}{0.5 \times 4} } = \sqrt{41} \approx 6.403$

The velocity of the particle at x = 10, v ≈ 6.403 m/s

(d) From the given values, the minimum value of the velocity is 5.22 m/s,

therefore, the velocity of the particle reduces from 6.0 m/s then increases,

but the particle does not stop or reverse to the origin.

The particle does not turn around and the particl does not head back

towards the origin.

(e) If the initial velocity of the particle at x = 0 is v = 2 m/s, we have;

Energy added to the particle;

$Work \ done \ on \ the \ particle, W = \dfrac{(4 \, m + 3 \, m) \times (-5 \, N)}{2} = -17.5 \, J$

Energy of the particle at point x = 0 is K.E. = 0.5×4×2² - 17.5 = -9.5

The energy added to the particle is more than the energy in the

particle, and given that the sign of the energy is negative, the particle

reverses direction, and the work is done in the opposite direction to the

initial direction of the particle.

Therefore;

Yes the particle turns around.

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A particle moving along the x-axis is acted upon by the force whose functional form appears below. T

answered: sindy35111

a) 5.57 m/s

b) 5.22 m/s

c) 6.40 m/s

d) No

e) Yes

Explanation:

The graph of the force is missing: find it in attachment. Also, the mass of the particle is 4.0 kg.

From the work-energy theorem, the work done by the force on the particle is equal to the change in kinetic energy of the particle. Therefore, we can write:

$Fx=\frac{1}{2}mv^2-\frac{1}{2}mu^2$

where:

F is the force applied on the particle

x is the displacement of the particle

m = 4.0 kg is the mass of the particle

v is the final velocity of the particle

u = 6.0 m/s is the initial velocity of the particle

In this first part,

x = 2.0 m

F = -5.0 N (the force is constant)

Therefore, the final velocity is:

$v=\sqrt{u^2+\frac{2Fx}{m}}=\sqrt{6.0^2+\frac{2(-5.0)(2.0)}{4.0}}=5.57 m/s$

In this part, we must be careful because we cannot use the formula used in part a), since the force is not constant.

However, we can find the work done by calculating the area under the graph; so we have:

$W=\frac{1}{2}mv^2-\frac{1}{2}mu^2$

Where W, the work done, is the area under the graph.

At x = 4.0 m, the figure is a trapezoid with bases 3 and 4 and height -5, therefore the area is:

$W=\frac{(4+3)\cdot (-5)}{2}=-17.5 J$

Then we also have

u = 6.0 m/s

m = 4.0 kg

So, the final velocity is

$v=\sqrt{u^2+\frac{2W}{m}}=\sqrt{6.0^2+\frac{2(-17.5)}{4.0}}=5.22 m/s$

Here we want to find the velocity at x = 10.0 m.

Therefore, we have to calculate the area of the graph between x = 0 and and x = 4.0 m is identical (in magnitude) to the area between x = 4.0 and x = 8.0 m, since the two figures are specular: however, their sign are opposite, so they cancel each other. This means that the work done on the particle between x = 0 and x = 4 m is -17.5 J, while the work done on the particle between x = 4 m and x = 8 m is +17.5 J: so, the net work done on the particle between x =0 and x=8 m is zero.

Therefore, here we have to calculate only the work done on the particle between x = 8.0 m and x = 10.0 m, which is:

W=Fx=(5.0 N)(10.0 m-8.0 m)=10 J

And so, the final velocity is:

$v=\sqrt{u^2+\frac{2W}{m}}=\sqrt{6.0^2+\frac{2(10)}{4.0}}=6.40 m/s$

No, it doesn't.

We can answer this part by looking at the answers of part a, b and c.

At the beginning, the force exerted on the particle is negative: therefore, we see that the velocity of the particle decreases, because the work done on it is negative (so, the particle slows down. However, the minimum velocity is reached at x = 4.0 m, where it is v = 5.22 m/s. But the velocity is still positive: this means that the particle is still moving forward.

After this point, the force on the particle becomes positive: this means that it will start to "push" the particle, so the particle will start accelerating. Therefore, after this point, the particle will keep moving in the same direction with more and more velocity.

Here the initial velocity of the particle is

u = 2.0 m/s

To see if the particle turns back at some point, we have to study the situation in part b), at x = 4.0 m.

As we said in part b), the work done on the particle between x = 0 and x = 4.0 m is

W=-17.5 J

Therefore, the particle's velocity at x = 4.0 m is

$v=\sqrt{u^2+\frac{2W}{m}}=\sqrt{2.0^2+\frac{2(-17.5)}{4.0}}$

We see that the argument of the square root is negative: this means that the particle's velocity at this point is negative. So, this means that yes, the particle has turned back.