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yawa3891 [41]
3 years ago
11

A block of mass m=1kg sliding along a rough horizontal surface is traveling at a speed v0=2m/s when it strikes a massless spring

head-on and compresses the spring a maximum distance X=10cm. If the spring has stiffness constant k=10N/m, determine the coefficient of kinetic friction between block and surface.
Physics
1 answer:
topjm [15]3 years ago
4 0

Here we can use the work energy theorem

W_f + W_s = K_f - K_i

here we know that

K_f = 0

as it come to rest finally

K_i = \frac{1}{2}mv_i^2

K_i = \frac{1}{2}\times 1\times 2^2

K_i = 2 J

now work done by friction force will be given as

W_f = - F_f \times d = -\mu mg d

W_f = - \mu(1)(9.8)(0.10) = - 0.98\mu

Work done by spring force is given as

W_s = \frac{1}{2}k(x_i^2 - x_f^2)

W_s = \frac{1}{2}(10)( 0 - 0.10^2)

W_s = -0.05 J

so now plug in all data above

- 0.05 - \mu(0.98) = 0 - 2

\mu = 1.99

so above is the friction coefficient


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Answer:

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500 C = 773 K

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p2 = 35*0.1*773/(293*1.3) = 7.1 bar

The index of expansion then is 35/7.1 = 4.93

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3 years ago
A 0.200-m uniform bar has a mass of 0.795 kg and is released from rest in the vertical position, as the drawing indicates. The s
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Explanation:

Since, the rod is present in vertical position and the spring is unrestrained.

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Initial total energy T = \frac{mgL}{2}

Now, when the rod is in horizontal position then final total energy will be as follows.

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where,    I = moment of inertia of the rod about the end = \frac{mL^{2}}{3}

Also,    \omega = \frac{\nu}{L}

where,    \nu = speed of the tip of the rod

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The initial unstrained length is x_{o} = 0.1 m

Therefore, final length will be calculated as follows.

              x' = \sqrt{(0.2)^{2} + (0.1)^{2}} m

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Putting the given values into the above formula as follows.

   \frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1}{6}mv^{2}

  \frac{0.795 kg \times 9.8 \times 0.2 m}{2} = \frac{1}{2} \times 27 N/m \times (0.1236)^{2} + \frac{1}{6} \times 0.795 \times v^{2}

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Anna71 [15]

Answer:

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