3. 8.0 MJ of work is inputted to isothermally move a 1500 Kg mass from an elevation of zero to 35m and to change it velocity. The initial velocity of the mass is 0.55 m/s. The drag work for this process is described by a force which varies over the distance travel,F = ax + b; where F is in Newtons, the coefficient (a) has units of N/m and the constant (b) has units of N.For this application a = 4.1 N/m and b = 700 N. The force is applied over a distance of 1730 m. There is a heat loss from friction of 1.1(106) J during this process. Determine the final velocity of the mass.3b. [15] The input work for this process is from a combustion engine that has an efficiency of 0.28. The heating value of the fuel is 30.0 MJ/liters and its cost is $0.83/liter. Determine the volume of fuel required and its cost to perform the process described in (3.).

Question
Asked Oct 26, 2019

3. 8.0 MJ of work is inputted to isothermally move a 1500 Kg mass from an elevation of zero to 35m and to change it velocity. The initial velocity of the mass is 0.55 m/s. The drag work for this process is described by a force which varies over the distance travel,
F = ax + b; where F is in Newtons, the coefficient (a) has units of N/m and the constant (b) has units of N.
For this application a = 4.1 N/m and b = 700 N. The force is applied over a distance of 1730 m. There is a heat loss from friction of 1.1(106) J during this process. Determine the final velocity of the mass.
3b. [15] The input work for this process is from a combustion engine that has an efficiency of 0.28. The heating value of the fuel is 30.0 MJ/liters and its cost is $0.83/liter. Determine the volume of fuel required and its cost to perform the process described in (3.).

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Step 1

Given data as per question

Mass m 1500kg
Initial velocity,u 0.55 m/s
Gain in height 35-0-35m
10m/s2
F-ax+b
where a 4.1 N/m
b-700 N
total distance moved S=173m
work done by friction w, 1.1x10 j
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Mass m 1500kg Initial velocity,u 0.55 m/s Gain in height 35-0-35m 10m/s2 F-ax+b where a 4.1 N/m b-700 N total distance moved S=173m work done by friction w, 1.1x10 j

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Step 2
using work energy theoram
net
Wr W au+aKE
au change in potential energy
change in kinetic energy
AKE
Fd -1.1x10mgah+mv -;mv (1)
1730
Fdc (ax+bdx
1730
(1730)
+bx1730
2
W
+bx
= a
a-
F
2
W 1730[ax 865 + b]
W =1730[4.1x 865+ 7001j
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using work energy theoram net Wr W au+aKE au change in potential energy change in kinetic energy AKE Fd -1.1x10mgah+mv -;mv (1) 1730 Fdc (ax+bdx 1730 (1730) +bx1730 2 W +bx = a a- F 2 W 1730[ax 865 + b] W =1730[4.1x 865+ 7001j

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Step 3

Using (1) and (2) an...

w2 7.3x10j(2)
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w2 7.3x10j(2)

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