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N76 [4]
3 years ago
5

A certain spring is found not to obey Hooke’s law; it exerts a restoring force Fx(x)=−αx−βx2 if it is stretched or compressed, w

here α=60.0N/m and β=18.0N/m2. The mass of the spring is negligible. (a) Calculate the potential-energy function U(x) for this spring. Let U = 0 when x = 0. (b) An object with mass 0.900 kg on a frictionless, horizontal surface is attached to this spring, pulled a distance 1.00 m to the right (the +x-direction) to stretch the spring, and released. What is the speed of the object when it is 0.50 m to the right of the x = 0 equilibrium position?
Physics
1 answer:
Masteriza [31]3 years ago
4 0

Answer:U(x) = 30x^2 +6x^3

V^2=8.28m/s

Explanation:The law of conservation of energy is given by K1+U1= K2+U2 ...eq 1

Kinetic energy K.E= 1/2 mv^2

Restoring force function F(x)= -60x - 18x^2

But F(x)= -dU/dx

dU(x)=-F(x)dx

Integrating U(x)= -integral F(x)dx + U(0)

Substituting, we get

U(x) = - integral(-60x-18x^2)dx+U(0)

U(x)= 30x^2+6x^3+U(0)

U=0 at x=0

Therefore U(x)= 30x^2+6x^3

b) Given : x1=1.00m,x2= 0.50m ,V1=0, V2=?

Substituting into eq (a)

U1= 30(1.00)^2+6(1.00)^3=36J

Using x2=0.5 into eq(a)

U2=30(0.50)^2+6(0.50)^3=8.25J

Object at rest K1=0

0+36=K2+8.25

K2=27.75J

Given; m =0.900kg, V2=?

27.75=1/2×0.900×V2^2

V2= SQRT(2×27.75)/0.81

V2= 8.28m/s

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A block of mass 0.1 kg is attached to a spring of spring constant 21 N/m on a frictionless track. The block moves in simple harm
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Answer:

A) 2.75 m/s  B) 0.1911 m    C) 0.109 s

Explanation:

mass of block = M =0.1 kg

spring constant = k = 21 N/m

amplitude = A = 0.19 m

mass of bullet = m = 1.45 g = 0.00145 kg

velocity of bullet = vᵇ = 68 m/s

as we know:

Angular frequency of S.H.M = ω₀ = \sqrt\frac{k}{M}

                                                       = \sqrt\frac{21}{0.1}

                                                       = 14.49 rad/sec

<h3>A) Speed of the block immediately before the collision:</h3>

displacement of Simple Harmonic  Motion is given as:

                                x = A sin (\omega t + \phi)\\

Differentiating this to find speed of the block immediately before the collision:

                    v=\frac{dx}{dt}= A\omega_{o} cos (\omega_{o}t =\phi}\\

As bullet strikes at equilibrium position so,

                                  φ = 0

                                   t= 2nπ

                             ⇒ cos (ω₀t + φ) = 1

                             ⇒ v= A\omega_{o}

                                       v=(.19)(14.49)\\v= 2.75 ms^{-1}

<h3>B) If the simple harmonic motion after the collision is described by x = B sin(ωt + φ), new amplitude B:</h3>

S.H.M after collision is given as :

                              x= Bsin(\omega t + \phi)

To find B, consider law of conservation of energy

K.E = P.E\\K.E= \frac{1}{2}(m+M)v^{2}  \\P.E = \frac{1}{2} kB^{2}

\frac{m+M}{k} v^{2} = B^{2} \\B =\sqrt\frac{m+M}{k} v\\B = \sqrt\frac{.00145+0.1}{21} (2.75)\\B = .1911m

<h3>C) Time taken by the block to reach maximum amplitude after the collision:</h3>

Time period S.H.M is given as:

T=2\pi \sqrt\frac{m}{k}\\ for given case\\m= m=M\\then\\T=2\pi \sqrt\frac{m+M}{k}

Collision occurred at equilibrium position so time taken by block to reach maximum amplitude is equal to one fourth of total time period

T=\frac{\pi }{2}\sqrt\frac{m+M}{k} \\T=0.109 sec

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