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Math  /  Algebra

QuestionA 2.00 kg block (mass 1 ) and a 4.00 kg block (mass 2 ) are connected by a light string as shown; the inclination of the ramp is 40.0∘40.0^{\circ}. Friction is negligib What is (a) the acceleration of each block and (b) the tension in the string?

Studdy Solution

STEP 1

1. The system consists of two blocks connected by a light, inextensible string.
2. The mass of block 1 is 2.00 kg.
3. The mass of block 2 is 4.00 kg.
4. The ramp is inclined at an angle of 40.0∘40.0^{\circ}.
5. Friction is negligible.
6. We need to find (a) the acceleration of each block, and (b) the tension in the string.

STEP 2

1. Analyze the forces acting on each block.
2. Apply Newton's second law to each block.
3. Solve the equations to find the acceleration.
4. Use the acceleration to find the tension in the string.

STEP 3

Analyze the forces acting on each block.
For block 1 (2.00 kg), the forces are: - Gravitational force: Fg1=m1⋅g F_{g1} = m_1 \cdot g - Tension in the string: T T
For block 2 (4.00 kg), the forces are: - Gravitational force: Fg2=m2⋅g F_{g2} = m_2 \cdot g - Component of gravitational force along the incline: Fparallel=m2⋅g⋅sin⁡(40.0∘) F_{\text{parallel}} = m_2 \cdot g \cdot \sin(40.0^{\circ}) - Tension in the string: T T

STEP 4

Apply Newton's second law to each block.
For block 1: m1⋅a=T−m1⋅g m_1 \cdot a = T - m_1 \cdot g
For block 2: m2⋅a=m2⋅g⋅sin⁡(40.0∘)−T m_2 \cdot a = m_2 \cdot g \cdot \sin(40.0^{\circ}) - T

STEP 5

Solve the equations to find the acceleration.
Add the two equations to eliminate T T :
m1⋅a+m2⋅a=m2⋅g⋅sin⁡(40.0∘)−m1⋅g m_1 \cdot a + m_2 \cdot a = m_2 \cdot g \cdot \sin(40.0^{\circ}) - m_1 \cdot g
(m1+m2)⋅a=m2⋅g⋅sin⁡(40.0∘)−m1⋅g (m_1 + m_2) \cdot a = m_2 \cdot g \cdot \sin(40.0^{\circ}) - m_1 \cdot g
a=m2⋅g⋅sin⁡(40.0∘)−m1⋅gm1+m2 a = \frac{m_2 \cdot g \cdot \sin(40.0^{\circ}) - m_1 \cdot g}{m_1 + m_2}
Substitute the known values: a=4.00⋅9.81⋅sin⁡(40.0∘)−2.00⋅9.812.00+4.00 a = \frac{4.00 \cdot 9.81 \cdot \sin(40.0^{\circ}) - 2.00 \cdot 9.81}{2.00 + 4.00}
Calculate a a : a=4.00⋅9.81⋅0.6428−2.00⋅9.816.00 a = \frac{4.00 \cdot 9.81 \cdot 0.6428 - 2.00 \cdot 9.81}{6.00}
a=25.17−19.626.00 a = \frac{25.17 - 19.62}{6.00}
a=5.556.00 a = \frac{5.55}{6.00}
a=0.925 m/s2 a = 0.925 \, \text{m/s}^2

STEP 6

Use the acceleration to find the tension in the string.
Substitute a a into the equation for block 1:
T=m1⋅a+m1⋅g T = m_1 \cdot a + m_1 \cdot g
T=2.00⋅0.925+2.00⋅9.81 T = 2.00 \cdot 0.925 + 2.00 \cdot 9.81
T=1.85+19.62 T = 1.85 + 19.62
T=21.47 N T = 21.47 \, \text{N}
The acceleration of each block is 0.925 m/s2 0.925 \, \text{m/s}^2 and the tension in the string is 21.47 N 21.47 \, \text{N} .

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