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2 years ago

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A. A cord passing over an easily turned pulley (one that is both massless and frictionless) has 7kg mass hanging from one end an

d a 9kg mass hanging from the other end as shown below. G 7.0 kg 9.0 ks This arrangement is known as Atwood's machine. Calculate the (i). acceleration of the masses (ii). tension in the cord. [2.5

1 answer:

Zepler [3.9K]2 years ago

6 0

Answer:

M g = M a = a (9 - 7) kg with the two masses going in different directions

(7 + 9) kg * g = a * 2 kg 2 kg effective in prducing acceleration

a = 16 / 2 * g = 9.8 m/s^2 / 8 = 1.23 m/s^2

M1 a = T - M1 * g T producing upward acceleration of 1.23 m/s^2

T = 7 (a + g) = 7 * 11 = 77 N

Check:

M2 a = M2 * g - T M2 producing downward acceleration

T = M2 (g - a) = 9 * (9.8 - 1.23) = 77 N

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A 1 kg mass is attached to a spring with spring constant 7 Nt/m. What is the frequency of the simple harmonic motion? What is th

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3. 0.4 m

The amplitude in a simple harmonic motion corresponds to the maximum displacement of the mass-spring system. In this case, the mass is initially displaced by 0.4 m: this means that during its oscillation later, the displacement cannot be larger than this value (otherwise energy conservation would be violated). Therefore, this represents the maximum displacement of the mass-spring system, so it corresponds to the amplitude.

4. 0.19 m

We can solve this part of the problem by using the law of conservation of energy. In fact:

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$E=K=\frac{1}{2}mv^2$

where m = 1 kg and v = 0.5 m/s is the initial velocity of the mass

- When the spring reaches the maximum compression/stretching (x=A=amplitude), the velocity of the system is zero, so the kinetic energy is zero, and all the mechanical energy is just elastic potential energy:

$E=U=\frac{1}{2}kA^2$

Since the total energy must be conserved, we have:

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5. Amplitude of the motion: 0.44 m

We can use again the law of conservation of energy.

- $E_i = \frac{1}{2}kx_0^2 + \frac{1}{2}mv_0^2$ is the initial mechanical energy of the system, with $x_0=0.4 m$ being the initial displacement of the mass and $v_0=0.5 m/s$ being the initial velocity

$- E_f = \frac{1}{2}kA^2$ is the mechanical energy of the system when x=A (maximum displacement)

Equalizing the two expressions, we can solve to find A, the amplitude:

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Equalizing the two expressions, we can solve to find $v_{max}$ , the maximum velocity:

$\frac{1}{2}kx_0^2 + \frac{1}{2}mv_0^2=\frac{1}{2}mv_{max}^2\\v_{max}=\sqrt{\frac{k}{m}x_0^2+v_0^2}=\sqrt{\frac{7 N/m}{1 kg}(0.4 m)^2+(0.5 m/s)^2}=1.17 m/s m$

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3 years ago

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Nostrana [21]

Answer:

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