Answer:
A. This option is true
B. This option is false
C. This option is true
Explanation:
A. Generally speaking, the greater percentage of cold, the recrystallization grain size would turn out to be smaller. Therefore this true.
B. A higher annealing temperature does not result in smaller recrystallization grain size. Therefore this is false.
C. As the percentage of cold work is greater, the recrystallization temperature would tend to be lower. Therefore this is true.
Problema:
Una nevera de vinos, con un peso bruto de 50 kg., que tiene las siguientes dimensiones: .60 m Largo x .49 m ancho x .50 m altura. Para ser transportadas en un contenedor de 40 pies D.V. responder las siguientes preguntas:
• 1.Cuántas neveras de vinos de acuerdo al volumen caben en un contenedor de 40 pies?
• De acuerdo dimensiones internas (largo, ancho y alto), ¿Cuántas caben en un contenedor de 40 pies?
• De acuerdo al peso que soporta el contenedor. ¿Cuántas neveras de vinos es posible transportar?
Answer:
I can't understand this language .
Determine the pressure difference in N/m2,between two points 800m apart in horizontal pipe-line,150 mm diameter, discharging water at the rate of 12.5litres per second. Take the frictional coefficient ,f, as being 0.008
Answer: [tex]10.631\times 10^3\ N/m^2[/tex]
Explanation:
Given
Discharge is [tex]Q=12.5\ L[/tex]
Diameter of pipe [tex]d=150\ mm[/tex]
Distance between two ends of pipe [tex]L=800\ m[/tex]
friction factor [tex]f=0.008[/tex]
Average velocity is given by
[tex]\Rightarrow v_{avg}=\dfrac{12.5\times 10^{-3}}{\frac{\pi }{4}(0.15)^2}\\\\\Rightarrow v_{avg}=\dfrac{15.9134\times 10^{-3}}{2.25\times 10^{-2}}\\\\\Rightarrow v_{avg}=7.07\times 10^{-1}\\\Rightarrow v_{avg}=0.707\ m/s[/tex]
Pressure difference is given by
[tex]\Rightarrow \Delta P=f\ \dfrac{L}{d}\dfrac{\rho v_{avg}^2}{2}\\\\\Rightarrow \Delta P=0.008\times \dfrac{800}{0.15}\times \dfrac{997\times (0.707)^2}{2}\\\\\Rightarrow \Delta P=10,631.45\ N/m^2\\\Rightarrow \Delta P=10.631\ kPa[/tex]
Steam enters an adiabatic turbine at 6 MPa, 600°C, and 80 m/s and leaves at 50 kPa, 100°C, and 140 m/s. If the power output of the turbine is 5 MW, determine (a) the reversible power output and (b) the second-law efficiency of the turbine. Assume the surroundings to be at 25°C.
Answer:
(a) the reversible power output of turbine is 5810 kw
(b) The second-law efficiency of he turbine = 86.05%
Explanation:
In state 1: the steam has a pressure of 6 MPa and 600°C. Obtain the enthalpy and entropy at this state.
h1 = 3658 kJ/kg s1=7.167 kJ/kgK
In state 2: the steam has a pressure of 50 kPa and 100°C. Obtain the enthalpy and entropy at this state
h2 = 2682kl/kg S2= 7.694 kJ/kg
Assuming that the energy balance equation given
Wout=m [h1-h2+(v1²-v2²) /2]
Let
W =5 MW
V1= 80 m/s V2= 140 m/s
h1 = 3658kJ/kg h2 = 2682 kJ/kg
∴5 MW x1000 kW/ 1 MW =m [(3658-2682)+ ((80m/s)²-(140m/s)²)/2](1N /1kg m/ s²) *(1KJ/1000 Nm)
m = 5.158kg/s
Consider the energy balance equation given
Wrev,out =Wout-mT0(s1-s2)
Substitute Wout =5 MW m = 5.158kg/s 7
s1= 7.167 kJ/kg-K s2= 7.694kJ/kg-K and 25°C .
Wrev,out=(5 MW x 1000 kW /1 MW) -5.158x(273+25) Kx(7.167-7.694)
= 5810 kW
(a) Therefore, the reversible power output of turbine is 5810 kw.
The given values of quantities were substituted and the reversible power output are calculated.
(b) Calculating the second law efficiency of the turbine:
η=Wout/W rev,out
Let Wout = 5 MW and Wrev,out = 5810 kW
η=(5 MW x 1000 kW)/(1 MW *5810)
η= 86.05%
Activity 1. Fill the blank with the correct answer. Write your answer on the blank. 1. ___________________ is a regular pattern of dots displayed on the screen which acts as a visual aid and also used to define the extent of your drawing. 2. Ortho is short for ___________________, which means either vertical or horizontal. 3. Tangent is a point where two _______________________ meet at just a single point. 4. If you want to create a new drawing, simply press ___________________ for the short cut key. 5. There are _______________Osnap that can help you performs your task easier.
Answer:
1. Drawing grid.
2. Orthogonal.
3. Geometries.
4. CTRL+N.
5. Thirteen (13).
Explanation:
CAD is an acronym for computer aided design and it is typically used for designing the graphical representation of a building plan. An example of a computer aided design (CAD) software is AutoCAD.
Some of the features of an AutoCAD software are;
1. Drawing grid: is a regular pattern of dots displayed on the screen of an AutoCAD software, which acts as a visual aid and it's also used to define the extent of a drawing.
2. Ortho is short or an abbreviation for orthogonal, which means either vertical or horizontal.
3. Tangent is a point where two geometries meet at just a single point.
4. If you want to create a new drawing, simply press CTRL+N for the short cut key.
5. There are thirteen object snaps (Osnap) that can help you perform your task on AutoCAD easily. The 13 object snaps (Osnap) are; Endpoint, Midpoint, Apparent intersect, Intersection, Quadrant, Extension, Tangent, Center, Insert, Perpendicular, Node, Parallel, and Nearest.
what type of slab and beam used in construction of space neddle
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Consider the equation y = 10^(4x). Which of the following statements is true?
A plot of log(y) vs. x would be linear with a slope of 4.
A plot of log(y) vs. log (x) would be linear with a slope of 10.
A plot of log(y) vs. x would be linear with a slope of 10.
A plot of y vs. log(x) would be linear with a slope of 4.
A plot of log(y) vs. log (x) would be linear with a slope of 4.
A plot of y vs. log(x) would be linear with a slope of 10.
Answer: Plot of [tex]\log y[/tex] vs [tex]x[/tex] would be linear with a slope of 4.
Explanation:
Given
Equation is [tex]y=10^{4x}[/tex]
Taking log both sides
[tex]\Rightarrow \log y=4x\log (10)\\\Rightarrow \log y=4x[/tex]
It resembles with linear equation [tex]y=mx+c[/tex]
Here, slope of [tex]\log y[/tex] vs [tex]x[/tex] is 4.
Your shifts productivity is Slow because one person is not pulling his share. The rest of the team is Getting upset.
Answer:
you are right but then you ddnt ask a question
Set the leak rate to zero and choose a non-zero value for the proportional feedback gain.Restart the simulation and turn on the outflow valve.What happens to the liquid level in the tank?Repeat this process with higher and lower values for the proportional feedback gain.What happens when the proportional feedback gain is increased?What happens when it is decreased?Find the proportional gain that will reach steady state the quickest without oscillationin the state of the valve and restart the simulation.What is the system time constant, as determined from the tank level versus time plot.
Answer:
Explanation:
The proportional gain K is usually a fixed property of the controller . If proportional gain is increased , The sensitivity of the controller to error is increased but the stability is impaired. The system approaches the behaviour of on off controlled system and it response become oscillatory
Imagine a cantilever beam fixed at one end with a mass = m and a length = L. If this beam is subject to an inertial force and a uniformly distributed load = w, what is the moment present at a length of L/4?
Answer:
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Explanation:
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grudbAir is compressed in a well insulated compressor from 95 kPa and 27 C to 600 kPa and 277 C. Use the air tables; assume negligible changes in kinetic and potential energy. Find the isentropic efficiency of the compressor. Find the exit temperature of the air if the compressor was reversible.
Answer:
a) 1.9%
b) T2s = 505.5 k = 232.5°C
Explanation:
P1 = 95 kPa
T1 = 27°C = 300 k
P2 = 600 kPa
T1 = 277°c = 550 k
Table used : Table ( A - 17 ) Ideal gas properties of air
a) determining the isentropic efficiency of the compressor
Л = ( h2s - h1 ) / ( h2a - h1 ) ---- ( 1 )
where ; h1 = 300.19 kJ/kg , T1 = 300 K , h2a = 554.74 kJ/kg , T2 = 550 k
To get h2s we have to calculate the the value of Pr2 using Pr1(relative pressure)
Pr2 = P2/P1 * Pr = ( 600 / 95 ) * 1.306 hence; h2s = 500.72 kJ/kg
back to equation1
Л = 0.019 = 1.9%
b) Calculate the exit temperature of the air if compressor is reversible
if compressor is reversible the corresponding exit temperature
T2s = 505.5 k = 232.5°C
given that h2s = 500.72 kJ/kg
Find the general solution of the given differential equation. Give the largest interval over which the general solution is defined. Determine whether there are any transient terms in the general solution.
x dx/dy−y=x^2sinx
Answer:
Interval: x∈ ( 0, ∞ )
There are no transient terms
Explanation:
x (dy/dx) – y= x^2sinx
Attached below is the detailed solution of the Given problem
There are no transient terms found in the general solution
Interval: x∈ ( 0, ∞ )
Calculate density, specific weight and weight of one litter of petrol having specific gravity 0.7
Explanation:
mass=19kg
density=800kg/m³
volume=?
as we know that
density=mass/volume
density×volume=mass
volume=mass/density
putting the values
volume=19kg/800kg/m³
so volume=0.02375≈0.02m³
bending stress distribution is a.rectangle b.parabolic c.curve d.i section
The heat transfer surface area of a fin is equal to the sum of all surfaces of the fin exposed to the surrounding medium, including the surface area of the fin tip. Under what conditions can we neglect heat transfer from the fin tip?
Answer:
The explanation according to the given query is summarized in the explanation segment below.
Explanation:
If somehow the fin has become too lengthy, this same fin tip temperature approaches the temperature gradient and maybe we'll ignore heat transmission out from end tips.Additionally, effective heat transmission as well from the tip could be ignored unless the end tip surface is relatively tiny throughout comparison to its overall surface.Identify the first step in preparing a spectrophotometer for use.
A. Make sure all samples and the blank are ready for measurement.
B. Prepare a calibration curve.
C. Measure the absorbance of the blank.
D. Turn on the light source and the spectrophotometer.
Answer:
D. Turn on the light source and the spectrophotometer.
Explanation:
A spectrophotometer is a machine used to measure the presence of any light-absorbing particle in a solution as well as its concentration. To prepare a spectrophotometer for use, the first step is to turn on the spectrophotometer and allow it to warm up for at least 15 minutes. After this is done, the next step will be to ensure that the samples and blank are ready. Next, an appropriate wavelength is set for the solute being determined. Finally, the absorbance is measured of both the blank and samples.
The temperature gradient in a spherical (or cylindrical) wall at steady state will always decrease (in magnitude) with increasing distance from the center (line), i.e. radial distance.
A. True
B. False
Answer:
True
Explanation:
Yes it is true that the Temperature gradient would also decrease with magnitude just as the distances rise from the centre line.
We have this cylinder equation as
[T1-T2 / ln(r1-r2)]2πKL
The radial distance is r2-r1
The gradient of temperature is T1-T2
From the equation,
The temperature gradient has a direct and proportional relationship to radial distance
T1-T2 ∝ ln(r2-r1)
1/T1-T2 = k(r2-r1)
This inverse relationship above confirms that the statement is true
A flow inside a centrifuge can be approximated by a combination of a central cylinder and a radial line source flow, giving the following potential function:
Ø= a2/r -cosØ + aßlnr = r
Where a is the radius of the central base of the centrifuge and ß is a constant.
a) Provide expressions for the velocities Vr and vo .
b) Find the expression for the stream function.
Answer:
a) Vr = - a^2/r cosθ + aß / r
Vθ = 1/r [ -a^2/r * sinθ ]
b) attached below
Explanation:
potential function
Ø= a^2 /r cosØ + aßlnr ----- ( 1 )
a = radius , ß = constant
a) Expressions for Vr and Vθ
Vr = dØ / dr ----- ( 2 )
hence expression : Vr = - a^2/r cosθ + aß / r
Vθ = 1/r dØ / dθ ------ ( 3 )
back to equation 1
dØ / dr = - a^2/r sinθ + 0 --- ( 4 )
Resolving equations 3 and 4
Vθ = 1/r [ -a^2/r * sinθ ]
b) expression for stream function
attached below
An ideal neon sign transformer provides 9130 V at 51.0 mA with an input voltage of 240 V. Calculate the transformer's input power and current.
Answer:
Input power = 465.63 W
current = 1.94 A
Explanation:
we have the following data to answer this question
V = 9130
i = 0.051
the input power = VI
I = 51.0 mA = 0.051
= 9130 * 0.051
= 465.63 watts
the current = 465.63/240
= 1.94A
therefore the input power is 465.63 wwatts
while the current is 1.94A
the input power is the same thing as the output power.
Blocks A and B each have a mass m. Determine the largest horizontal force P which can be applied to B so that A will not move relative to B. All surfaces are smooth.
Answer:
The answer is "15 N".
Explanation:
Please find the complete question in the attached file.
In frame B:
For just slipping:
[tex]\to \frac{P}{2} \cos \theta =mg \sin \theta\\\\\to P=2 mg \tan \theta \\\\[/tex]
[tex]=2 \times 1 \times g \times \tan 37^{\circ}\\\\ =2 \times 10 \times \frac{3}{4}\\\\ =15 \ N[/tex]
The following is a correlation for the average Nusselt number for natural convection over spherical surface. As can be seen in the above, the Nusselt number approaches 2 as Rayleigh number approaches zero. Prove that this situation corresponds to conduction heat transfer and in conduction heat transfer over sphere, the Nusselt number becomes 2. Hint: First step: Write an expression for heat transfer between two spherical shells that share the same center. Second step: Assume the outer spherical shell is infinitely large.
Answer:
Explanation:
[tex]r_2=[/tex]∞
[tex]q=4\pi kT_1(T_2-T_1)\\[/tex]
[tex]q=2\pi kD.[/tex]ΔT--------(1)
[tex]q=hA[/tex] ΔT[tex]=4\pi r_1^2(T_2_s-T_1_s)\\[/tex]
[tex]N_u=\frac{hD}{k} = 2+\frac{0.589 R_a^\frac{1}{4} }{[1+(\frac{0.046}{p_r}\frac{9}{16} )^\frac{4}{9} } ------(3)[/tex]
By equation (1) and (2)
[tex]2\pi kD.[/tex]ΔT=h.4[tex]\pi r_1^2[/tex]ΔT
[tex]2kD=hD^2\\\frac{hD}{k} =2\\N_u=\frac{hD}{k}=2\\[/tex]-------(4)
From equation (3) and (4)
So for sphere [tex]R_a[/tex]→0
In low speed subsonic wind tunnels, the value of test section velocity can be controlled by adjusting the pressure difference between the inlet and test-section for a fixed ratio of inlet-to-test section cross-sectional area.
a. True
b. false
Answer:
Hence the given statement is false.
Explanation:
For low-speed subsonic wind tunnels, the air density remains nearly constant decreasing the cross-section area cause the flow to extend velocity, and reduce pressure. Similarly increasing the world cause to decrease and therefore the pressure to extend.
The speed within the test section is decided by the planning of the tunnel.
Thus by adjusting the pressure difference won't change the worth of test section velocity.
Answer:
The given statement is false .
A levee will be constructed to provide some flood protection for a residential area. The residences are willing to accept a one-in-five chance that the levee will be overtopped in the next 15 years. Assuming that the annual peak streamflow follows a lognormal distribution with a log10(Q[ft3/s]) mean and standard deviation of 1.835 and 0.65 respectively, what is the design flow in ft3/s?
Answer:
1709.07 ft^3/s
Explanation:
Annual peak streamflow = Log10(Q [ft^3/s] )
mean = 1.835
standard deviation = 0.65
Probability of levee been overtopped in the next 15 years = 1/5
Determine the design flow ins ft^3/s
P₁₅ = 1 - ( q )^15 = 1 - ( 1 - 1/T )^15 = 0.2
∴ T = 67.72 years
Q₁₅ = 1 - 0.2 = 0.8
Applying Lognormal distribution : Zt = mean + ( K₂ * std ) --- ( 1 )
K₂ = 2.054 + ( 67.72 - 50 ) / ( 100 - 50 ) * ( 2.326 - 2.054 )
= 2.1504
back to equation 1
Zt = 1.835 + ( 2.1504 * 0.65 ) = 3.23276
hence:
Log₁₀ ( Qt(ft^3/s) ) = Zt = 3.23276
hence ; Qt = 10^3.23276
= 1709.07 ft^3/s
Given resistance 30ohms Inductance 200mH is connected to a 230v,50hZ supply. Impedance 69.6ohms Calculate current consumed?
Answer:
the current consumed is 3.3 A
Explanation:
Given;
resistance, R = 30 ohms
inductance, L = 200 mH
Voltage supply, V = 230 V
frequency of the coil, f = 50 Hz
impedance, Z = 69.6 Ohms
The current consumed is calculated as;
[tex]I = \frac{V}{Z} \\\\I = \frac{230}{69.6} \\\\I = 3.3 \ A[/tex]
Therefore, the current consumed is 3.3 A
A 1m3 tank containing air at 25℃ and 500kPa is connected through a valve to
another tank containing 5kg of air at 35℃ and 200kPa. Now the valve is opened,
and the entire system is allowed to reach thermal equilibrium, which is at 20℃
(Take: Ru = 8.314 kJ / kg.K).
Answer:
The right answer is "2.2099 m³".
Explanation:
Given:
Mass,
m = 5 kg
Temperature,
T = 35℃
or,
= 35 + 273
Pressure,
P = 200 kPa
Gas constant,
R = 0.2870 kj/kgK
By using the ideal gas equation,
The volume will be:
⇒ [tex]PV=mRT[/tex]
or,
⇒ [tex]V=\frac{mRT}{P}[/tex]
By substituting the values, we get
[tex]=\frac{5(0.2870)(35+273)}{200}[/tex]
[tex]=\frac{441.98}{200}[/tex]
[tex]=2.2099 \ m^3[/tex]
A horizontal water jet impinges against a vertical flat plate at 30 ft/s and splashes off the sides in the verti- cal plane. If a horizontal force of 500 lbf is required to hold the plate against the water stream, determine the volume flow rate of the water.
Answer:
8.6 ft³/s
Explanation:
The force due to the water jet F = mv where m = mass flow rate = ρQ where ρ = density of water = 62.4 lbm/ft³ and Q = volume flow rate. v = velocity of water jet = 30 ft/s
So, F = mv
F = ρQv
making Q subject of the formula, we have
Q = F/ρv
Since F = force due to water jet = force needed to hold the plate against the water stream = 500 lbf = 500 × 1 lbf = 500 × 32.2 lbmft/s² = 16100 lbmft/s²
Since
Q = F/ρv
Substituting the values of the variables into the equation for Q, we have
Q = F/ρv
Q = 16100 lbmft/s²/(62.4 lbm/ft³ × 30 ft/s)
Q = 16100 lbmft/s²/1872 lbm/ft²s
Q = 8.6 ft³/s
So, the volume flow rate is 8.6 ft³/s.
Steam at 4 MPa and 350°C is expanded in an adiabatic turbine to 125kPa. What is the isentropic efficiency (percent) of this turbine if the steam is exhausted as a saturated vapor?
Answer:
[tex]\eta_{turbine} = 0.603 = 60.3\%[/tex]
Explanation:
First, we will find actual properties at given inlet and outlet states by the use of steam tables:
AT INLET:
At 4MPa and 350°C, from the superheated table:
h₁ = 3093.3 KJ/kg
s₁ = 6.5843 KJ/kg.K
AT OUTLET:
At P₂ = 125 KPa and steam is saturated in vapor state:
h₂ = [tex]h_{g\ at\ 125KPa}[/tex] = 2684.9 KJ/kg
Now, for the isentropic enthalpy, we have:
P₂ = 125 KPa and s₂ = s₁ = 6.5843 KJ/kg.K
Since s₂ is less than [tex]s_g[/tex] and greater than [tex]s_f[/tex] at 125 KPa. Therefore, the steam is in a saturated mixture state. So:
[tex]x = \frac{s_2-s_f}{s_{fg}} \\\\x = \frac{6.5843\ KJ/kg.K - 1.3741\ KJ/kg.K}{5.91\ KJ/kg.K}\\\\x = 0.88[/tex]
Now, we will find [tex]h_{2s}[/tex](enthalpy at the outlet for the isentropic process):
[tex]h_{2s} = h_{f\ at\ 125KPa}+xh_{fg\ at\ 125KPa}\\\\h_{2s} = 444.36\ KJ/kg + (0.88)(2240.6\ KJ/kg)\\h_{2s} = 2416.088\ KJ/kg[/tex]
Now, the isentropic efficiency of the turbine can be given as follows:
[tex]\eta_{turbine} = \frac{h_1-h_2}{h_1-h_{2s}}\\\\\eta_{turbine} = \frac{3093.3\ KJ/kg-2684.9\ KJ/kg}{3093.3\ KJ/kg-2416.088\ KJ/kg}\\\\\eta_{turbine} = \frac{408.4\ KJ/kg}{677.212\ KJ/kg}\\\\\eta_{turbine} = 0.603 = 60.3\%[/tex]
A 20-mm-diameter steel bar is to be used as a torsion spring. If the torsional stress in the bar is not to exceed 110 MPa when one end is twisted through an angle of 15°, what must be the length of the bar?
Answer:
1.887 m
Explanation:
(15 *pi)/180
= 0.2618 rad
Polar moment
= Pi*d⁴/32
= (22/7*20⁴)/32
= 15707.96
Torque on shaft
= ((22/7)*20³*110)/16
= 172857.14
= 172.8nm
Shear modulus
G = 79.3
L = Gjθ/T
= 79.3x10⁹x(1.571*10^-8)x0.2618/172.8
= 1.887 m
The length of the bar is therefore 1.887 meters
The propeller shaft of the submarine experiences both torsional and axial loads. Draw Mohr's Circle for a stress element on the outside surface of the solid shaft. Determine the principal stresses, the maximum in-plane shear stress and average normal stress using Mohr's Circle.
Answer: Attached below is the missing detail and Mohr's circle.
i) б1 = 9.6 Ksi
б2 = -10.7 ksi
ii) 10.2 Ksi
iii) -0.51Ksi
Explanation:
First step :
direct compressive stress on shaft
бd = P / π/4 * d^2
= -20 / 0.785 * 5^2 = -1.09 Ksi
shear stress at the outer surface due to torsion
ζ = 16*T / πd^3
= (16 * 250 ) / π * 5^3 = 010.19 Ksi
Calculate the Principal stress, maximum in-plane shear stress and average normal stress
Using Mohr's circle ( attached below )
i) principal stresses:
б1 = 4.8 cm * 2 = 9.6 Ksi
б2 = -5.35 cm * 2 = -10.7 ksi
ii) maximum in-plane shear stress
ζ = radius of Mohr's circle
= 5.1 cm = 10.2 Ksi ( Given that ; 1 cm = 2Ksi )
iii) average normal stress
= 9.6 + ( - 10.7 ) / 2
= -0.51Ksi
A work-mode-choice model is developed from data acquired in the field in order to determine the probabilities of individual travelers selecting various modes. The mode choices include automobile drive-alone (DL), automobile shared-ride (SR), and bus (B). The utility functions are estimated as follows:
UDL = 2.6 - 0.3 (costDL) - 0.02 (travel timeDL)
USR = 0.7 - 0.3 (costSR) - 0.04 (travel timeSR)
UB = -0.3 (costB) - 0.01 (travel timeB)
where cost is in dollars and time is in minutes. The cost of driving an automobile is exist5.50 with a travel time of 21 minutes, while the bus fare is exist1.25 with a travel time of 27 minutes. How many people will use the shared-ride mode from a community of 4500 workers, assuming the shared-ride option always consists of three individuals sharing costs equally?
a. 314
b. 828
c. 866
d. 2805
Answer:
b. 828
Explanation:
UDL = 2.6 - 0.3 [5.5] - 0.02 [ 21 ] = 0.53
USR = 0.7 - 0.3 [5.5 / 2 ] - 0.04 [ 21 ] = -0.69
UB = -0.3 [ 1.25 ] - 0.01 = -0.645
Psr = [tex]\frac{e^{-0.53} }{e^{-0.53} + e^{-0.69} + e^{-0.645} }[/tex]
Solving the equation we get 0.184.
Number of people who will take shared ride is:
0.184 * 4500 = 828 approximately.
The system is initially moving with the cable taut, the 15-kg block moving down the rough incline with a speed of 0.080 m/s, and the spring stretched 39 mm. By the method of this article, (a) determine the velocity v of the block after it has traveled 99 mm, and (b) calculate the distance d traveled by the block before it comes to rest.
Solution :
The spring is expanded by 2 times of the block when it moves down an inclined by x times.
Here, [tex]$x_1$[/tex] = 39 mm
[tex]x_2[/tex] = 225 mm
a). From the work energy principal,
Work forces = kinetic energy
[tex]$(mg \sin 50^\circ)\times \frac{99}{1000}-(\mu_k mg \cos 50^\circ) \times \frac{99}{1000} -\frac{1}{2}k(0.225^2 - 0.039^2)=\frac{1}{2}m(V^2_2-0.08^2)$[/tex]
[tex]$(112.6 \times 0.099)-(14.17 \times 0.099)-4.91= 7.5(V^2_2-0.08^2)$[/tex]
[tex]$9.75= 7.5(V^2_2-0.08^2)$[/tex]
[tex]$1.3= V^2_2-0.08^2$[/tex]
[tex]$V_2=1.14\ m/s$[/tex]
b). calculating the distance travelled by the block before it comes to rest.
Substitute the value of [tex]V_2[/tex] in (1),
[tex]$-(\mu_kmg \cos 50^\circ)x + (mg \sin 50^\circ)x-\frac{1}{2}k\left( ( 2x+0.039)^2 - 0.039^2\right)= -\frac{1}{2}m(0.08)^2$[/tex]
[tex]$-14.17x+112.6x - 100(4x^2+0.156x)=-0.048$[/tex]
[tex]$98.43x - 100(4x^2+0.156x)+0.048=0$[/tex]
[tex]$98.43x - 400x^2-15.6x+0.048=0$[/tex]
[tex]$82.83x - 400x^2+0.048=0$[/tex]
[tex]$ 400x^2- 82.83x-0.048=0$[/tex]
x = 0.20 m
