Mathematics, 02.09.2019 17:30 Renebebe1

The figure shows the potential energy of a 400g particle as it moves along the x-axis. suppose the particle's mechanical energy is 12 j.
where are the particle's turning points?
what is the particle's speed when it is at x= 2.0m?
what is the particle's maximum speed?
at what position or positions does this occur?

The figure shows the potential energy of a 400g particle as it moves along the x-axis. suppose the p

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Answer from: jordanlopezorosco16

1) The turning points are at x = 1, x = 2, x = 4.

2) At x = 2, P.E = 8 J
So KE = 12 - 8 = 4 J
4 = 1/2 x 0.4 x v²
v = 4.47 m/s

The maximum speed is when K.E = 12 J
12 = 1/2 x 0.4 x v²
v = 7.74 m/s

This occurs at the points where P.E is 0, so at x = 1 and x = 4

Answer from: jonzyjones3114

the particle's turning points : x = 1, x = 2, x = 4.

the particle's speed when it is at x= 2.0 : v = 4,472 m/s

the particle's maximum speed : v = 7,746 m/s, at x=1 and x=4

Further explanation

Energy is the ability to do work

Energy because its motion is expressed as Kinetic energy (KE) which can be formulated as:

$\large{\boxed{\bold{KE=\frac{1}{2}mv^2}}}$

So for two objects that have the same speed, the greater the mass of the object, the greater the kinetic energy

Whereas for a stationary object it has kinetic energy = 0

While the energy produced from its position is called potential energy (PE)

which can be formulated as:

Ep = m. g. h ⇒ gravitational potential energy

The sum of kinetic and potential energy is called mechanical energy

ME = KE + PE turning points

The particle's turning points is the point where the particle changes its direction of motion, because the motion of a particle can be considered as a vector

In the picture, turning points are shown at x = 1, x = 2, x = 4.

the particle's speed when it is at x = 2.0 m :

The potential energy (PE) value in the figure when x = 2 is 8 J

Because it is known that ME = 12 J, particle velocity can be found from the kinetic energy equation

KE = ME - PE

KE = 12 - 8

KE = 4 J

$\rm KE=\frac{1}{2}mv^2\\\\m=400\:g=0.4\:kg\\\\4=\frac{1}{2}0.4.v^2\\\\v=4,472\frac{m}{s}$

the particle's maximum speed

The maximum particle velocity occurs when PE = 0 and KE = ME

The lowest particle velocity occurs when KE = 0 and PE = ME

So :

KE = ME

$\rm KE=1\:J\\\\12=\frac{1}{2}.0.4.v^2\\\\v=7,746\frac{m}{s}$

The position when the particle on the maximum speed (PE = 0) occurs when particles at x = 1 and x = 4

Learn more

The kinetic energy of the vehicle

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law of motion

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displacement of A skateboarder

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Keywords: kinetic energy, speed,particle,turning points, motion, work, potential energy, mechanical energy

Answer from: Quest

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Answer from: Quest

x = ± 6i

step-by-step explanation:

x^2 = -36

take the square root of each side

sqrt(x^2) = sqrt(-36)

we know sqrt(ab) = sqrt(a) sqrt(b)

x = ±sqrt(36) sqrt(-1)

we know that sqrt (-1) = i

x = ± 6i

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