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3 years ago

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A block with mass m = 0.250 kg is attached to one end of an ideal spring and moves on a horizontal frictionless surface. The oth

er end of the spring is attached to a wall. When the block is at x = +0.240 m, its acceleration is ax = -13.0 m/s2 and its velocity is vx = +4.00 m/s.1. What is the spring's force constant k?2. What is the amplitude of the motion?3. What is the maximum speed of the block during its motion?4. What is the maximum magnitude of the block's acceleration during its motion?

1 answer:

ehidna [41]3 years ago

5 0

Answer:

13.54 N/m

0.6 m

4.37 m/s

32.496 m/s²

Explanation:

m = Mass of block = 0.25 kg

k = Spring constant

A = Amplitude

x = Compression of spring = 0.24 m

a = Acceleration = -13 m/s²

v = Velocity = 4 m/s

The weight of the block and force on spring is equal

$ma=-kx\\\Rightarrow k=-\frac{ma}{x}\\\Rightarrow k=-\frac{0.25\times -13}{0.24}\\\Rightarrow k=13.54\ N/m$

The spring's force constant is 13.54 N/m

Total energy of the system is given by

$E=\frac{1}{2}(mv^2+kx^2)\\\Rightarrow E=\frac{1}{2}(0.25\times 4^2+13.54\times 0.24^2)\\\Rightarrow E=2.39\ J$

At maximum displacement v = 0

$E=\frac{1}{2}(mv^2+kA^2)\\\Rightarrow E=\frac{1}{2}(0+kA^2)$

$\\\Rightarrow E=\frac{1}{2}kA^2\\\Rightarrow A=\sqrt{\frac{E2}{k}}\\\Rightarrow A=\sqrt{\frac{2\times 2.39}{13.54}}\\\Rightarrow A=0.6\ m$

The amplitude of the motion is 0.6 m

Speed of the block

$E=\frac{1}{2}mv_m^2+0\\\Rightarrow v_m=\sqrt{\frac{2E}{m}}\\\Rightarrow v_m=\sqrt{\frac{2\times 2.39}{0.25}}\\\Rightarrow v_m=4.37\ m/s$

The maximum speed of the block during its motion is 4.37 m/s

Forces in the spring

$ma_m=kA\\\Rightarrow a_m=\frac{kA}{m}\\\Rightarrow a_m=\frac{13.54\times 0.6}{0.25}\\\Rightarrow a_m=32.496\ m/s^2$

Maximum magnitude of the block's acceleration during its motion is 32.496 m/s²

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" $x(t)= e^\frac{-t}{2}((\frac{-4}{3})\cos\frac{\sqrt{47}}{2}t- \frac{-64\sqrt{47}}{141} \sin\frac{\sqrt{47}}{2}t)+\frac{10}{3}(\cos(3t)+ \sin (3t))$ ".

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Taking into consideration a volume weight = 16 pounds originally extends a springs $\frac{8}{3}$ feet but is extracted to resting at 2 feet beneath balance position.

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$W=mg\\m=\frac{w}{g}\\m=\frac{16}{32}\\m= \frac{1}{2} slug\\$

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$16= \frac{8}{3} k \\\\8k=16 \times 3 \\\\k=16\times \frac{3}{8} \\\\k=6 \frac{lb}{ft}\\\\$

Number $\frac{1}{2}$ times the immediate speed, i.e .. Damping force

$\frac{1}{2} \frac{d^2 x}{dt^2} = -6x-\frac{1}{2}\frac{dx}{dt}+10 \cos 3t \\\\\frac{1}{2} \frac{d^2 x}{dt^2}+ \frac{1}{2}\frac{dx}{dt}+6x =10 \cos 3t \\ \\\frac{d^2 x}{dt^2} +\frac{dx}{dt}+12x=20\cos 3t \\\\$

The m^2+m+12=0 and m is an auxiliary equation,

$m=\frac{-1 \pm \sqrt{1-4(12)}}{2}\\\\m=\frac{-1 \pm \sqrt{47i}}{2}\\\\\ m1= \frac{-1 + \sqrt{47i}}{2} \ \ \ \ or\ \ \ \ \ m2 =\frac{-1 - \sqrt{47i}}{2}$

Therefore, additional feature

$x_c (t) = e^{\frac{-t}{2}}[C_1 \cos \frac{\sqrt{47}}{2}t+ C_2 \sin \frac{\sqrt{47}}{2}t]$

Use the form of uncertain coefficients to find a particular solution.

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These values are replaced by equation ( 1):

$\frac{d^2x}{dt}+\frac{dx}{dt}+ 12x=20 \cos(3t) -9 Acos(3t) -9B sin (3t) -3Asin(3t)+3B cos (3t) + 12A cos (3t) + 12B sin (3t)\\\\3Acos 3t + 3B sin 3t - 3Asin 3t + 3B cos 3t= 20cos(3t)\\(3A+3B)cos3t -(3A-3B)sin3t = 20 cos (3t)\\$

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$3B -3A=O\\3(\frac{10}{3})-3A =0\\A= \frac{10}{3}\\$

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Initially, the volume of rest x(0)=2 and x'(0) is extracted by rest i.e.

Throughout the general solution, replace initial state x(0) = 2,

Replace x'(0)=0 with a general solution in the initial condition,

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3 years ago

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Explanation:

The acceleration of an object given it's velocity and time taken can be found by using the formula

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where

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t is the time taken

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Since the body is from rest u = 0

From the question we have

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We have the final answer as

<h3>4.6 m/s²</h3>

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