Two blocks connected by a string are pulled across a horizontal surface by a force applied to one of the blocks, as shown. The coefficient of kinetic friction between the blocks and the surface is 0.25. If each block has an acceleration of 2.0m/s2 to the right, what is the magnitude F of the applied force?
The Correct Answer and Explanation is :
To determine the magnitude of the applied force ( F ) required to pull two blocks connected by a string across a horizontal surface with a given acceleration, we need to consider the forces acting on the system, including friction and the applied force’s components.
Given Data:
- Mass of block 1 (( m_1 )): 1.0 kg
- Mass of block 2 (( m_2 )): 3.0 kg
- Coefficient of kinetic friction (( \mu_k )): 0.25
- Acceleration of the system (( a )): 2.0 m/s²
- Angle of applied force (( \theta )): 60°
Free-Body Diagrams:
- Block 1 (1.0 kg):
- Forces:
- Tension (( T )) to the right
- Frictional force (( f_1 )) to the left
- Normal force (( N_1 )) upward
- Gravitational force (( W_1 = m_1 \cdot g )) downward
- Block 2 (3.0 kg):
- Forces:
- Applied force (( F )) at an angle ( \theta = 60° )
- Tension (( T )) to the left
- Frictional force (( f_2 )) to the left
- Normal force (( N_2 )) upward
- Gravitational force (( W_2 = m_2 \cdot g )) downward
Calculations:
- Determine Normal Forces:
- For block 1:
[
N_1 = W_1 = m_1 \cdot g = 1.0\, \text{kg} \times 9.8\, \text{m/s}^2 = 9.8\, \text{N}
] - For block 2, considering the vertical component of the applied force:
[
N_2 = W_2 – F \cdot \sin(\theta) = m_2 \cdot g – F \cdot \sin(60°)
]
[
N_2 = 3.0\, \text{kg} \times 9.8\, \text{m/s}^2 – F \cdot \frac{\sqrt{3}}{2} = 29.4\, \text{N} – 0.866F
]
- Calculate Frictional Forces:
- Frictional force is given by ( f = \mu_k \cdot N ).
- For block 1:
[
f_1 = \mu_k \cdot N_1 = 0.25 \times 9.8\, \text{N} = 2.45\, \text{N}
] - For block 2:
[
f_2 = \mu_k \cdot N_2 = 0.25 \times (29.4\, \text{N} – 0.866F) = 7.35\, \text{N} – 0.2165F
]
- Apply Newton’s Second Law:
- For block 1:
[
T – f_1 = m_1 \cdot a
]
[
T – 2.45\, \text{N} = 1.0\, \text{kg} \times 2.0\, \text{m/s}^2
]
[
T = 4.45\, \text{N}
] - For block 2:
[
F \cdot \cos(60°) – T – f_2 = m_2 \cdot a
]
[
0.5F – 4.45\, \text{N} – (7.35\, \text{N} – 0.2165F) = 3.0\, \text{kg} \times 2.0\, \text{m/s}^2
]
[
0.5F – 4.45\, \text{N} – 7.35\, \text{N} + 0.2165F = 6.0\, \text{N}
]
[
0.7165F – 11.8\, \text{N} = 6.0\, \text{N}
]
[
0.7165F = 17.8\, \text{N}
]
[
F = \frac{17.8\, \text{N}}{0.7165} \approx 24.85\, \text{N}
]
Conclusion:
The magnitude of the applied force ( F ) required to move the two blocks with an acceleration of 2.0 m/s², considering the given coefficient of kinetic friction, is approximately 24.85 N.