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2 years 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:

Blizzard [7]2 years ago

7 0

Answer:

The potential energy is $-4x^3$

Explanation:

Given that,

Force $F=-\alpha x^2 i$

We need to calculate the potential energy

Using formula of work done

$\Delta U=F(x) dx$

Put the value of F

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

On integration

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

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

U = 0, x = 0 so C = 0

Put the value of c and α in equation (I)

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

$U=-4x^3$

Hence, The potential energy is $-4x^3$

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<h3>PROOF:</h3>

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∑Fₓ = Maₓ

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$f_{s} - kx = Ma_{x}$ [/tex] -----(8)

Using (6) in (8)

$\frac{1}{2}Ma_{x} - kx =Ma_{x}$

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$\omega^{2} = \frac{2k}{3M}$

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A 100 cm3 block of lead weighs 11N is carefully submerged in water. One cm3 of water weighs 0.0098 N.

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#1

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V = 11.22 cm^3

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#2

buoyant force on the lead block is balancing the weight of it

$F_B = W$

$\rho V g = W$

$13* 10^3 * V * 9.8 = 11.11$

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(iii) Buoyant force = 11.11 N

(iv) since the density of lead is less than the density of mercury so it will float inside mercury

#3

Yes, if object density is less than the density of liquid then it will float otherwise it will sink inside the liquid

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2 years ago