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4 months ago

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In an experiment, one of the forces exerted on a proton is F⃗ =−αx2i^, where α=12N/m2. What is the potential-energy function for

F⃗ ? Let U=0 when x=0.

1 answer:

serg [3.5K]4 months ago

7 0

Answer:

The potential energy is given by $-4x^3$ .

Explanation:

In relation to the force,

Force is represented by $F=-\alpha x^2 i$ .

We will now calculate the potential energy

Employing the work done formula

$\Delta U=F(x) dx$

Substituting the value of F results in:

$\Delta U=-\alpha x^2\ dx$

Upon integration, we derive:

$U=-\alpha \dfrac{x^3}{3}+C$

$U=-\dfrac{\alpha x^3}{3}+C$ ...(I)

Setting U = 0 at x = 0 gives C = 0.

Substitute the values of c and α into equation (I):

$U=-\dfrac{12x^3}{3}+0$

$U=-4x^3$

Therefore, the potential energy is $-4x^3$ .

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Read 2 more answers

Two projectiles are launched with the same initial speed from the same location, one at a 30° angle and the other at a 60° angle

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Answer:

Explanation:

Let us denote the launch angle as $\theta _1=30^{\circ}$ .

$\theta _2=60^{\circ}$ , $\theta _1$ , and $\theta _2$ are complementary angles which means their ranges are identical.

$H_{max}=\frac{u^2(\sin \theta )^2}{2g}$

$H_{30}=\frac{u^2(\sin 30)^2}{2g}$

$H_{30}=\frac{u^2}{8g}$

Time of flight is indicated as $=\frac{2u\sin \theta }{g}$ .

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$H_{60}=\frac{u^2(\sin 60)^2}{2g}$

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$H_{60}=3\times H_{30}$

$t_{60}=\sqrt{3}\times t_{30}$

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