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Montano1993 [528]
3 years ago
7

A 1.2 kg block is held at rest against the spring with a force constant k= 730 N/m. Initially, the spring is compressed a distan

ce d. When the block is released, it slides across a surface that is frictionless except for a rough patch of width 5.0 cm that has a coefficient of kinetic friction = 0.44. Find d, such that the block's speed after crossing the rough patch is 2.3 m/s.
Physics
1 answer:
mariarad [96]3 years ago
5 0

Answer:

d = 9.69 cm

Explanation:

given,

mass of the block = 1.2 Kg

spring force constant(k) = 730 N/m

spring is compressed = d = ?

rough patch width = 5 cm

μ_k = 0.44

work done by friction = energy lost

\mu_k mg \times x = \dfrac{1}{2}kd^2-\dfrac{1}{2}mv^2

0.44\times 1.2 \times 9.8 \times 0.05 = \dfrac{1}{2}\times 730 \times d^2-\dfrac{1}{2}\times 1.2 \times 2.3^2

3.432 = 365 d^2

d = \dfrac{3.432}{365}

d = 0.0969 m

d = 9.69 cm

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Answer:

Flow Rate = 80 m^3 /hours  (Rounded to the nearest whole number)

Explanation:

Given

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Calculations

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The energy equation for this system will be,

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The other three equations to solve the above equations are:

Re = (rho*V*D)/ μ

Flow Rate, Q = V*(pi/4)*D^2

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Re = 235000

f = 0.015

V = 1.97 m/s

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Q = V*(pi/4)*D^2

Q = (1.97)*(pi/4)*(0.12)^2 = 0.022 m^3/s = 80 m^3 /hours

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