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Tju [1.3M]
3 years ago
11

A cubic shaped box has a side length of 1.0 ft and a mass of 10 lbm is sliding on a frictionless horizontal surface towards a 30

upward incline. The horizontal velocity of the box is 20 ft/s. Determine how far up the incline the box will travel (report center of mass distance along the inclined surface, not vertical distance)
Engineering
1 answer:
natulia [17]3 years ago
6 0

Explanation  & answer:

Assuming a smooth transition so that there is no abrupt change in slopes to avoid frictional loss nor toppling, we can use energy considerations.

Initially, the cube has a kinetic energy of

KE = mv^2/2 = 10 lbm * 20^2 ft^2/s^2  / 2 = 2000 lbm-ft^2 / s^2

At the highest point when the block stops, the gain in potential energy is

PE = mgh = 10 lbm * 32.2 ft/s^2 * h ft = 322 lbm ft^2/s^2

By assumption, there was no loss in energies, we equate PE = KE

322h lbm ft^2/s^2 = 2000 lbm ft^2/s^2

=>

h = 2000 /322 = 6.211 (ft)

distance up incline = h / sin(30) = 12.4 ft

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Oil with a density of 800 kg/m3 is pumped from a pressure of 0.6 bar to a pressure of 1.4 bar, and the outlet is 3 m above the i
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Q = Volumetric flow rate = 0.2 m³/s

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Velocity through outlet = V₂ = Q/A₂ = 0.2/0.03 = 6.67 m/s

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\frac {P_1}{\rho g}+\frac{V_1^2}{2g}+z_1+h=\frac {P_2}{\rho g}+\frac{V_2^2}{2g}+z_2\\\Rightarrow h=\frac{P_2-P_1}{\rho g}+\frac{V_2^2-V_1^2}{2g}+z_2-z_1\\\Rightarrow h=\frac{(1.4-0.6)\times 10^5}{800\times 9.81}+\frac{6.67_2^2-3.33^2}{2\times 9.81}+3\\\Rightarrow h=14.896\ m

Work done by pump

W_{p}=\rho gQh\\\Rightarrow W_{p}=800\times 9.81\times 0.2\times 14.896\\\Rightarrow W_{p}=23380.8\ W

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h=\frac{P_2-P_1}{\rho g}+z_2-z_1\\\Righarrow h=\frac{(1.4-0.6)\times 10^5}{800\times 9.81}+3\\\Righarrow h=13.19\ m\\

Work done by pump

W_{p}=\rho gQh\\\Rightarrow W_{p}=800\times 9.81\times 0.2\times 13.193\\\Rightarrow W_{p}=20708.8\ W

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