**10 Dynamics Object connected by cord over pulley Atwood machine – Problems and Solutions**

1. Block A with a mass of 5 kg, placed on a smooth horizontal plane. Block B with a mass of 3 kg hanging at one end of the cord connected with block A over a pulley. Acceleration due to gravity is 10 m/s^{2}. What is the acceleration of both blocks?

__Known :__

Mass of block A (m_{A}) = 5 kg

Mass of block B (m_{B}) = 3 kg

Acceleration due to gravity (g) = 10 m/s^{2}

Weight of block B (w_{B}) = m_{B} g = (3)(10) = 30 Newton

__Wanted:__ Acceleration of both blocks (a)

__Solution :__

The horizontal plane is smooth so there is no friction force. The force that accelerates both blocks is the weight of the block B.

ΣF = m a

w_{B} = (m_{A} + m_{B}) a

30 = (5 + 3) a

30 = 8 a

a = 30 / 8

a = 3.75 m/s^{2}

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2.

Based on the figure above,

(1) acceleration of object = 0

(2) the object moves at a constant velocity

(3) object at rest

(4) the object moves if the weight of the object is smaller than the force that pulls the object.

Solution :

(1) The acceleration of the object = 0.

*Net force :*

∑*F = m a –> **acceleration **(a) = 0*

∑*F = 0*

*F*_{1}* + F*_{2}* – F*_{3}* = 12 + 24 – 36 = 36 – 36 = 0 N*

(2) The object moves at a constant velocity

*No acceleration means object at rest or moves at a constant velocity. *

(3) object at rest

*No net force means object at rest.*

(4) the object moves if the weight of the object is smaller than the force that pulls the object.

*Weight acts on the vertical direction, while the pull force acts on the horizontal direction.*

*The object moves in a horizontal direction so only the horizontal forces that act on the object*

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3. If the coefficient of kinetic friction between the block A and the table surface is 0.1. Acceleration due to gravity is 10 m/s2, then what is the force acts on the block A so that the system moves to leftward in 2 m/s^{2}.

__Known :__

Mass of block A (m_{A}) = 30 kg

*weight of block **A (w*_{A}*) = (30 kg)(10 m/s*^{2}*) = 300 kg m/s*^{2}* **or **300 Newton*

Mass of block B (m_{B}) = 20 kg

*weight of block **B (w*_{B}*) = (20 kg)(10 m/s*^{2}*) = 200 kg m/s*^{2 }*or **200 Newton*

Acceleration due to gravity (g) = 10 m/s^{2}

Coefficient of kinetic friction (μ_{k}) = 0.1

Acceleration of system (a) = 2 m/s^{2} (to leftward)

*Force of kinetic friction **(f*_{k}*) = **μ*_{k }*N = **μ*_{k }*w*_{A}* = (0.1)(300) = 30 Newton *

__Wanted :__ The magnitude of force F

Solution :

Newton’s second law :

ΣF = m a

The object A moves to leftward :

F – f_{k} – w_{B} = (m_{A} + m_{B}) a

F – 30 – 200 = (30 + 20)(2)

F – 230 = (50)(2)

F – 230 = 100

F = 230 + 100

F = 330 Newton

4. Two objects, A = 2 kg and B = 6 kg, attached at one end of cord over a pulley, as shown in figure below. If acceleration due to gravity is 10 ms^{-2} then what is the acceleration of the object B.

__Known :__

Mass of object A (m_{A}) = 2 kg, m_{B} = 6 kg, g = 10 m/s^{2}

weight of object A (w_{A}) = (m_{A})(g) = (2)(10) = 20 N

weight of object B (w_{B}) = (m_{B})(g) = (6)(10) = 60 N

__Wanted :__ Acceleration of object b (system’s acceleration).

__Solution :__

w_{B} > w_{A} so that object B moves downward, object A moves upward

ΣF = m a

w_{B} – w_{A} = (m_{A} + m_{B}) a

60 – 20 = (2 + 6) a

40 = (8) a

a = 5 m/s^{2}

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5. Two objects connected by a cord over a smooth pulley, as shown in figure below. If m_{1} = 1 kg, m_{2} = 2 kg, and acceleration due to gravity is 10 ms^{-2}, then what is the tension force T.

__Known :__

Mass of object 1 (m_{1}) = 1 kg

Mass of object 2 (m_{2}) = 2 kg

Acceleration due to gravity (g) = 10 m/s^{2}

weight of object 1 (w_{1}) = m_{1} g = (1 kg)(10 m/s^{2}) = 10 kg m/s^{2} or 10 Newton

weight of object 2 (w_{2}) = m_{2 }g = (2 kg)(10 m/s^{2}) = 20 kg m/s^{2} or 20 Newton

__Wanted :__ The tension force (T) ?

__Solution :__

w_{2} > w_{1 }so m_{2} moves downward, m_{1} moves upward.

The Newton’s second law of motion :

ΣF = m a

w_{2} – w_{1} = (m_{1} + m_{2}) a

20 – 10 = (1 + 2 ) a

10 = (3) a

a = 3.3 m/s^{2}

System’s acceleration = 3.3 m/s^{2}.

m_{2} moves downward :

w_{2} – T_{2} = m_{2} a

20 – T_{2} = (2)(3.33)

20 – T_{2} = 6.66

T_{2} = 20 – 6.66

T_{2} = 13.3 Newton

m_{1} moves upward :

T_{1} – w_{1} = m_{1} a

T_{1} – 10 = (1)(3.3)

T_{1} – 10 = 3.33

T_{1} = 10 + 3.33

T_{1} = 13.3 Newton

The tension force (T) = 13.3 Newton.

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6. Mass of m_{1} = 6 kg and mass of m_{2} = 4 kg. The horizontal surface is smooth. Acceleration due to gravity is 10 m/s^{2}. What is the system’s acceleration.

__Known :__

Mass of m_{1} = 6 kg

mass of m_{2 }= 4 kg

acceleration due to gravity (g) = 10 m/s^{2}

weight of w_{1 }= m_{1 }g = (6 kg)(10 m/s^{2}) = 60 kg m/s^{2} or 60 Newton

weight of w_{2} = m_{2} g = (4 kg)(10 m/s^{2}) = 40 kg m/s^{2} or 40 Newton

__Wanted :__ System’s acceleration (a)

Solution :

m_{1} on a smooth horizontal plane without friction so that the system accelerated by the weight of the block 2.

Apply Newton’s second law :

∑F = m a

w_{2} = (m_{1} + m_{2}) a

40 N = (6 kg + 4 kg) a

40 N = (10 kg) a

a = 40 N / 10 kg

a = 4 m/s^{2}

7. Two blocks, each block has the mass of 2 kg, connected by a cord over a pulley, as shown in the figure below. The horizontal plane and pulley are smooth. If the block B is pulled by a horizontal force of 40 Newton, then what is the acceleration of block. Acceleration due to gravity is 10 m/s^{2}.

__Known :__

mass of block A (m_{A}) = mass of block B (m_{B}) = 2 kg

Acceleration due to gravity (g) = 10 m/s^{2}

Force of F = 40 N

weight of object A (w_{A}) = m g = (2)(10) = 20 N

__Wanted :__ System’s acceleration (a) ?

__Solution :__

Apply Newton’s second law :

∑F = m a

F – w_{A} = (m_{A} + m_{B}) a

40 – 20 = (2 + 2) a

20 = (4) a

a = 20 / 4

a = 5 m/s^{2}

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8. Mass of block A = 2 kg and mas of block B = 1 kg. Block B initially at rest, then accelerated downward until it hits ground. Acceleration due to gravity is 10 m/s^{2}. What is the magnitude of the tension force.

__Known :__

Mass of block A (m_{A}) = 2 kg

Mass of block B (m_{B}) = 1 kg

Acceleration due to gravity (g) = 10 m/s^{2}

Weight of block B (w_{B}) = m_{B} g = (1)(10) = 10 Newton

__Wanted :__ The magnitude of the tension force (T)

__Solution :__

Ignore the friction force.

__System’s acceleration (a)__

∑F = m a

w_{B} = (m_{A }+ m_{B}) a

10 = (2 + 1) a

10 = 3 a

a = 10/3

__The tension force ____(T)__

The tension force on the block A :

∑F = m a

T = m_{A} a = (2)(10/3) = 20/3 = 6.7 Newton

__The tension force on the block B :__

∑F = m a

w_{B} – T = m_{B} a

10 – T = (1)(10/3)

10 – T = 3.3

T = 10 – 3.3 = 6.7 Newton

The tension force (T) = 6.7 Newton

9. Mass of block A = 2 kg and mass of block B = 1 kg. The friction force between object A with the horizontal plane = 2.5 Newton. Ignore friction on pulley and cord. What is the acceleration of both blocks.

__Known :__

Mass of block A (m_{A}) = 2 kg

Mass of block B (m_{B}) = 1 kg

Friction force between lock a and the horizontal plane (f_{kA}) = 2.5 Newton

Acceleration due to gravity (g) = 10 m/s^{2}

Weight of block (w_{B}) = m_{B} g = (1)(10) = 10 Newton

__Wanted :__ Acceleration of both blocks (a)

__Solution :__

Apply Newton’s second law :

∑F = m a

w_{B} – f_{k} = (m_{A} + m_{B}) a

10 – 2.5 = (2 + 1) a

7.5 = 3 a

a = 7.5 / 3

a = 2.5 m/s^{2}

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10. Mass of block a = 30 kg, rest on a horizontal plane connected with block B with mass of 10 kg over a pulley. What is the system’s acceleration. Acceleration due to gravity is 10 ms^{-2}.

__Known :__

Mass of block A (m_{A}) = 30 kg

Mass of block B (m_{B}) = 10 kg

Acceleration due to gravity (g) = 10 m/s^{2}

weight of block B (w_{B}) = m_{B} g = (10)(10) = 100 Newton

__Wanted :__ System’s acceleration (a)

__Solution :__

∑F = m a

w_{B} = (m_{A} + m_{B}) a

100 = (30 + 10) a

100 = 40 a

a = 100 / 40

a = 2.5 m/s^{2}