Physics: The two blocks are attached to each other by a massless string that is wrapped around a frictionless pulley as shown in figure-2. When the bottom 4.00 kg block is pulled to the left by the constant force P~ = 45 N, the top 2.00 kg block slides across it to the right. Assume that the coefficient of kinetic friction between all surfaces is μk = 0.400

Free fall without air resistance:g=9.81 m/s²t=235 sh=?h=0.5*g*t²h=0.5*9.81*235²h=270879m =270 km...

The two blocks are attached to each other by a massless

Ответ:

holmesleauja

10.02.2022

a) The complete free body diagram of the system is presented in the image attached below according to the Newton's Law.

b) The acceleration of the top block is 0.962 meters per square second in the -x direction and the acceleration of the bottom block is 0.962 meters per square second in the -x direction.

c) The tension on the rope is 9.770 newtons.

Explanation:

a) The complete free body diagram of the system is presented in the image attached below according to the Newton's Law.

b) At first we have to construct corresponding equations of equilibrium for each mass:

Mass A

$\Sigma F_{x} = T - \mu_{k}\cdot N_{A} = m_{A}\cdot a_{A}$ (1)

$\Sigma F_{y} = N_{A}-m_{A}\cdot g = 0$ (2)

Mass B

$\Sigma F_{x} = -P+\mu_{k}\cdot N_{A}+\mu_{k}\cdot N_{B}+T= -m_{B}\cdot a_{A}$ (3)

$\Sigma F_{y} = N_{B}-N_{A}-m_{B}\cdot g= 0$ (4)

Where:

- Force exerted on the bottom block, measured in newtons.

$N_{A}$ , $N_{B}$ - Normal forces, measured in newtons.

$m_{A}$ , $m_{B}$ - Masses of the top and bottom blocks, measured in kilograms.

$\mu_{k}$ - Kinetic coeffcient of friction, dimensionless.

- Gravitational acceleration, measured in meters per square second.

$a_{A}$ , $a_{B}$ - Net accelerations of the top and bottom blocks, measured in meters per square second.

From (2):

$N_{A} = m_{A}\cdot g$ (5)

(2) in (4):

$N_{B}=(m_{A}+m_{B})\cdot g$ (6)

(5) in (1):

$T = m_{A}\cdot a_{A}+\mu_{k}\cdot m_{A}\cdot g$ (7)

(6) in (3):

$T = -m_{B}\cdot a_{A}+P-\mu_{k}\cdot m_{A}\cdot g-\mu_{k}\cdot (m_{B}+m_{A})\cdot g$

$T = -m_{B}\cdot a_{A}+P -\mu_{k}\cdot (2\cdot m_{A}+m_{B})\cdot g$ (8)

And we obtain an expression for the magnitude of the acceleration by eliminating tension:

$m_{A}\cdot a_{A} +\mu_{k}\cdot m_{A}\cdot g = -m_{B}\cdot a_{A}+P-\mu_{k}\cdot (2\cdot m_{A}+m_{B})\cdot g$

$(m_{A}+m_{B})\cdot a_{A} = P-\mu_{k}\cdot (3\cdot m_{A}+m_{B})\cdot g$

$a_{A} = \frac{P-\mu_{k}\cdot (3\cdot m_{A}+m_{B})\cdot g}{m_{A}+m_{B}}$

If we know that $P = 45\,N$ , $m_{A} = 2\,kg$ , $m_{B} = 4\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ and $\mu_{k} = 0.4$ , then the magnitude of the acceleration is:

$a_{A} = \frac{45\,N-(0.4)\cdot [3\cdot (2\,kg)+4\,kg]\cdot \left(9.807\,\frac{m}{s^{2}} \right)}{2\,kg+4\,kg}$

$a_{A} = 0.962\,\frac{m}{s^{2}}$

The acceleration of the top block is 0.962 meters per square second in the -x direction and the acceleration of the bottom block is 0.962 meters per square second in the -x direction.

c) By means of (7) and knowing that $m_{A} = 2\,kg$ , $a_{A} = 0.962\,\frac{m}{s^{2}}$ , $\mu_{k} = 0.4$ and $g = 9.807\,\frac{m}{s^{2}}$ , we find that the tension on the rope is:

$T = (2\,kg)\cdot \left(0.962\,\frac{m}{s^{2}} \right)+(0.4)\cdot (2\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)$

$T = 9.770\,N$

The tension on the rope is 9.770 newtons.

The two blocks are attached to each other by a massless string that is wrapped

around a frictionles

0,0(0 оценок)

Ответ:

Ответ:

Ask your question to the AI advisor

More tips

Answers on questions: Physics

Ask an AI advisor a question