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/s2. What is the acceleration of both blocks?

Known :

Mass of block A (mA) = 5 kgDynamics, object connected by cord over pulley, atwood machine - problems and solutions 1

Mass of block B (mB) = 3 kg

Acceleration due to gravity (g) = 10 m/s2

Weight of block B (wB) = mB 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

wB = (mA + mB) a

30 = (5 + 3) a

30 = 8 a

a = 30 / 8

a = 3.75 m/s2

Read :  Determine horizontal displacement of projectile motion

2.

Based on the figure above, Dynamics, object connected by cord over pulley, atwood machine - problems and solutions 2

(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

F1 + F2 – F3 = 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

Read :  Parabolic motion, work and kinetic energy, linear momentum, linear and angular motion – problems and solutions

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/s2.

Known :

Mass of block A (mA) = 30 kgDynamics, object connected by cord over pulley, atwood machine - problems and solutions 3

weight of block A (wA) = (30 kg)(10 m/s2) = 300 kg m/s2 or 300 Newton

Mass of block B (mB) = 20 kg

weight of block B (wB) = (20 kg)(10 m/s2) = 200 kg m/s2 or 200 Newton

Acceleration due to gravity (g) = 10 m/s2

Coefficient of kinetic friction (μk) = 0.1

Acceleration of system (a) = 2 m/s2 (to leftward)

Force of kinetic friction (fk) = μk N = μk wA = (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 – fk – wB = (mA + mB) 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 (mA) = 2 kg, mB = 6 kg, g = 10 m/s2Dynamics, object connected by cord over pulley, atwood machine - problems and solutions 4

weight of object A (wA) = (mA)(g) = (2)(10) = 20 N

weight of object B (wB) = (mB)(g) = (6)(10) = 60 N

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

Solution :

wB > wA so that object B moves downward, object A moves upward

ΣF = m a

wB – wA = (mA + mB) a

60 – 20 = (2 + 6) a

40 = (8) a

a = 5 m/s2

Read :  Mechanical waves (frequency, period, wavelength, the wave speed) - problems and solutions

5. Two objects connected by a cord over a smooth pulley, as shown in figure below. If m1 = 1 kg, m2 = 2 kg, and acceleration due to gravity is 10 ms-2, then what is the tension force T.

Known :

Mass of object 1 (m1) = 1 kgDynamics, object connected by cord over pulley, atwood machine - problems and solutions 5

Mass of object 2 (m2) = 2 kg

Acceleration due to gravity (g) = 10 m/s2

weight of object 1 (w1) = m1 g = (1 kg)(10 m/s2) = 10 kg m/s2 or 10 Newton

weight of object 2 (w2) = m2 g = (2 kg)(10 m/s2) = 20 kg m/s2 or 20 Newton

Wanted : The tension force (T) ?

Solution :

w2 > w1 so m2 moves downward, m1 moves upward.

The Newton’s second law of motion :

ΣF = m a

w2 – w1 = (m1 + m2) a

20 – 10 = (1 + 2 ) a

10 = (3) a

a = 3.3 m/s2

System’s acceleration = 3.3 m/s2.

m2 moves downward :

w2 – T2 = m2 a

20 – T2 = (2)(3.33)

20 – T2 = 6.66

T2 = 20 – 6.66

T2 = 13.3 Newton

m1 moves upward :

T1 – w1 = m1 a

T1 – 10 = (1)(3.3)

T1 – 10 = 3.33

T1 = 10 + 3.33

T1 = 13.3 Newton

The tension force (T) = 13.3 Newton.

Read :  Newton's laws of motion – problems and solutions

6. Mass of m1 = 6 kg and mass of m2 = 4 kg. The horizontal surface is smooth. Acceleration due to gravity is 10 m/s2. What is the system’s acceleration.

Known :

Mass of m1 = 6 kgDynamics, object connected by cord over pulley, atwood machine - problems and solutions 6

mass of m2 = 4 kg

acceleration due to gravity (g) = 10 m/s2

weight of w1 = m1 g = (6 kg)(10 m/s2) = 60 kg m/s2 or 60 Newton

weight of w2 = m2 g = (4 kg)(10 m/s2) = 40 kg m/s2 or 40 Newton

Wanted : System’s acceleration (a)

Solution :

m1 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

w2 = (m1 + m2) a

40 N = (6 kg + 4 kg) a

40 N = (10 kg) a

a = 40 N / 10 kg

a = 4 m/s2

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/s2.

Known :

mass of block A (mA) = mass of block B (mB) = 2 kgDynamics, object connected by cord over pulley, atwood machine - problems and solutions 7

Acceleration due to gravity (g) = 10 m/s2

Force of F = 40 N

weight of object A (wA) = m g = (2)(10) = 20 N

Wanted : System’s acceleration (a) ?

Solution :

Apply Newton’s second law :

F = m a

F – wA = (mA + mB) a

40 – 20 = (2 + 2) a

20 = (4) a

a = 20 / 4

a = 5 m/s2

Read :  Double slit interference – problems and solutions

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/s2. What is the magnitude of the tension force.

Known :

Mass of block A (mA) = 2 kgDynamics, object connected by cord over pulley, atwood machine - problems and solutions 8

Mass of block B (mB) = 1 kg

Acceleration due to gravity (g) = 10 m/s2

Weight of block B (wB) = mB 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

wB = (mA + mB) 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 = mA a = (2)(10/3) = 20/3 = 6.7 Newton

The tension force on the block B :

F = m a

wB – T = mB 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 (mA) = 2 kgDynamics, object connected by cord over pulley, atwood machine - problems and solutions 9

Mass of block B (mB) = 1 kg

Friction force between lock a and the horizontal plane (fkA) = 2.5 Newton

Acceleration due to gravity (g) = 10 m/s2

Weight of block (wB) = mB g = (1)(10) = 10 Newton

Wanted : Acceleration of both blocks (a)

Solution :

Apply Newton’s second law :

F = m a

wB – fk = (mA + mB) a

10 – 2.5 = (2 + 1) a

7.5 = 3 a

a = 7.5 / 3

a = 2.5 m/s2

Read :  Centripetal force in uniform circular motion – problems and solutions

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 (mA) = 30 kgDynamics, object connected by cord over pulley, atwood machine - problems and solutions 10

Mass of block B (mB) = 10 kg

Acceleration due to gravity (g) = 10 m/s2

weight of block B (wB) = mB g = (10)(10) = 100 Newton

Wanted : System’s acceleration (a)

Solution :

F = m a

wB = (mA + mB) a

100 = (30 + 10) a

100 = 40 a

a = 100 / 40

a = 2.5 m/s2