Questions: If m1=23 kg and m2=38 kg, calculate the tension in the rope. 2 points 372.78 N 764.7 N 143.9 N 281.2 N If m1=27 kg and m2=438 kg, what is the acceleration of the system? 2 points

If m1=23 kg and m2=38 kg, calculate the tension in the rope.
2 points
372.78 N
764.7 N
143.9 N
281.2 N

If m1=27 kg and m2=438 kg, what is the acceleration of the system?
2 points

Transcript text: If $\mathrm{m} 1=23 \mathrm{~kg}$ and $\mathrm{m} 2=38 \mathrm{~kg}$, calculate the tension in the rope. * 2 points 372.78 N 764.7 N 143.9 N 281.2 N If $\mathrm{m} 1=27 \mathrm{~kg}$ and $\mathrm{m} 2=438 \mathrm{~kg}$, what is the acceleration of the system? * 2 points

Solution

Solution Steps

Step 1: Analyze the forces acting on each mass

For m1, the forces are tension (T) upwards and weight (m1_g) downwards. For m2, the forces are tension (T) upwards and weight (m2_g) downwards. Since m2 > m1, the system will accelerate downwards on the m2 side and upwards on the m1 side.

Step 2: Set up the equations of motion

For m1: T - m1_g = m1_a For m2: m2_g - T = m2_a

where 'g' is the acceleration due to gravity (approximately 9.8 m/s²) and 'a' is the acceleration of the system.

Step 3: Solve for acceleration (first question)

Given m1 = 27 kg and m2 = 438 kg, we can add the two equations from Step 2:

(T - m1_g) + (m2_g - T) = m1_a + m2_a m2_g - m1_g = (m1 + m2)_a a = (m2_g - m1*g) / (m1 + m2) a = ((438 kg * 9.8 m/s²) - (27 kg * 9.8 m/s²)) / (27 kg + 438 kg) a = 8.84 m/s²

Final Answer

8.84 m/s²

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