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Energy, Work, Power
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Energy, Work, Power

(a) Determine the work and power associated with a force of 7.5 x 10^-4 N acting over a distance of 2 meters in an elapsed time of 14 seconds
(b) Determine the work and power required to move a 5 kg mass up a frictionless plane inclined at an angle of 30^o with the horizontal for a distance of 2 m along the plane in a time of 3.5 seconds
(c) Determine the potential difference between point A and point B if 136 Joules of work is expended in moving
8.5 x 10¹⁸ electrons from point A to point B
(d) Determine the power and energy represented by a 305 V, 0.15 A pulse of electricity that lasts for 500 μs.
(e) Determine the amount of energy delivered in 2 hours by a 5 hp motor
(f) Determine the total energy transferred during t ≥ 0 by a circuit element with the following current and voltage values: current i = 10e^-5000t A; voltage v = 50(1 - e^-5000t) V
(g) Determine the power associated with a circuit element with the following charge profile and voltage value:
q = (4 x 10^-4)(1 - e^-250t) C for t ≥ 0; V = 10 V.

The strings: S₇P₃A₃₂ (force - push).
S₇P₃A₃₁ (force - pull).

The math:
Pj Problem of Interest is of type force (force - pull and push).

(a) Work = Force x Distance
So, Work = 7.5 x 10^-4 x 2 = 15 x 10^-4 = 1.5 mJ
Power is rate of doing work = Work/Time = (1.5 x 10^-3)/14 = 0.107 mW

(b) Force = mgcos60; where m is mass, g is acceleration due to gravity
So, Work = mgcos60 x 2 = 5 x 9.8 x 1/2 x 2 = 49 J
Power = 49/3.5 = 14 W.

(c) Voltage (V) = Work (W)/Charge (Q)
Charge per electron = 1.602 x 10^-19 C
So, Charge of 8.5 x 10¹⁸ electrons = 1.602 x 10^-19 x 8.5 x 10¹⁸ = 1.36 C
So, V = W/Q = 136/1.36 = 100 V.

(d) Electric Power (P) = Current (I) x Voltage (V)
So, P = 0.15(305) = 45.75 W
Electric Energy = Power x Time = IVt
So, Energy = 45.75 x 500 x 10^-6 = 22.9 mJ

(e) I horse power (hp) = 746 watts
So, 5 hp = 5 x 746 watts. 2 hours = 2 x 3600 seconds
So, Energy delivered by motor = 5 x 746 x 2 x 3600 = 26.9 MJ

(f) Total energy transferred = [ ∫ vidt ]₀^∞
So, Total energy transferred = 500[ (e^-5000t/-5000) + (e^-10000t/10000)]₀^∞ = 500/10000 = 50 mJ

(g) Current i = dq/dt = d[(4x10^-4)(1 - e^-250t)]/dt = (4x10^-4)(250e^-250t)
So, at t = 3 ms, i = 47.2 mA
So, power = 47.2 x 10 x 10^-3 = 472 mW

Math

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