Jay fills a wagon with sand (about 20 kg) and pulls it with a rope 40 m along the beach. He holds the rope 25â above the horizon
1 answer:
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Answer:
(a-1) d₂=4.89 m: The object slides 4.89 m along the rough surface
(a-2) Work (Wf) done by the friction force while the mass is sliding down the in- clined plane:
Wf= -20.4 J is negative
(b) Work (Wg) done by the gravitational force while the mass is sliding down the inclined plane:
Wg= 58.8 J is positive
Explanation:
Nomenclature
vf: final velocity
v₀ :initial velocity
a: acceleleration
d: distance
Ff: Friction force
W: weight
m:mass
g: acceleration due to gravity
Graphic attached
The attached graph describes the variables related to the kinetics of the object (forces and accelerations)
Calculation de of the components of W in the inclined plane
W=m*g
Wx₁ = m*g*sin30°
Wy₁= m*g*cos30°
Object kinematics on the inclined plane
vf₁²=v₀₁²+2*a₁*d₁
v₀₁=0
vf₁²=2*a₁*d₁
Equation (1)
Object kinetics on the inclined plane (μ= 0.2)
∑Fx₁=ma₁ :Newton's second law
-Ff₁+Wx₁ = ma₁ , Ff₁=μN₁
-μ₁N₁+Wx₁ = ma₁ Equation (2)
∑Fy₁=0 : Newton's first law
N₁-Wy₁= 0
N₁- m*g*cos30°=0
N₁ = m*g*cos30°
We replace N₁ = m*g*cos30 and Wx₁ = m*g*sin30° in the equation (2)
-μ₁m*g*cos30₁+m*g*sin30° = ma₁ : We divide by m
-μ₁*g*cos30°+g*sin30° = a₁
g*(-μ₁*cos30°+sin30°) = a₁
a₁ =9.8(-0.2*cos30°+sin30°)=3.2 m/s²
We replace a₁ =3.2 m/s² and d₁= 3m in the equation (1)
Rough surface kinematics
vf₂²=v₀₂²+2*a₂*d₂ v₀₂=vf₁=4.38 m/s
0 =4.38²+2*a₂*d₂ Equation (3)
Rough surface kinetics (μ= 0.3)
∑Fx₂=ma₂ :Newton's second law
-Ff₂=ma₂
--μ₂*N₂ = ma₂ Equation (4)
∑Fy₂= 0 :Newton's first law
N₂-W=0
N₂=W=m*g
We replace N₂=m*g inthe equation (4)
--μ₂*m*g = ma₂ We divide by m
--μ₂*g = a₂
a₂ =-0.2*9.8= -1.96m/s²
We replace a₂ = -1.96m/s² in the equation (3)
0 =4.38²+2*-1.96*d₂
3.92*d₂ = 4.38²
d₂=4.38²/3.92
d₂=4.38²/3.92
(a-1) d₂=4.89 m: The object slides 4.89 m along the rough surface
(a-2) Work (Wf) done by the friction force while the mass is sliding down the in- clined plane:
Wf = - Ff₁*d₁
Ff₁= μ₁N₁= μ₁*m*g*cos30°= -0.2*4*9.8*cos30° = 6,79 N
Wf= - 6.79*3 = 20.4 N*m
Wf= -20.4 J is negative
(b) Work (Wg) done by the gravitational force while the mass is sliding down the inclined plane
Wg=W₁x*d= m*g*sin30*3=4*9.8*0.5*3= 58.8 N*m
Wg= 58.8 J is positive
Hey there!
In this case, it is possible to solve this problem by using the widely-known steam tables which show that at 90 °C, the pressure that produces a vapor-liquid mixture at equilibrium is about 70.183 kPa (Cengel, Thermodynamics 5th edition).
Moreover, for the calculation of the volume, it is necessary to calculate the volume of the vapor-liquid mixture, given the quality (x) it has:
Thus, since 8 kg correspond to liquid water, 2 kg must correspond to steam, so that the quality turns out:
Now, at this temperature and pressure, the volume of a saturated vapor is 2.3593 m³/kg whereas that of the saturated liquid is 0.001036 m³/kg and therefore, the volume of the mixture is:
This means that the volume of the container will be:
Regards!