To determine the force required to accelerate the entire train at \(3 , \text{m/s}^2\), we first need to calculate the total mass of the train.
The lead car has a mass of \(2000 , \text{kg}\) and the other two cars each have a mass of \(1500 , \text{kg}\).
Calculating the total mass:
\[ \text{Total Mass} = \text{Mass of Lead Car} + \text{Mass of 2nd Car} + \text{Mass of 3rd Car} \]
\[ \text{Total Mass} = 2000 , \text{kg} + 1500 , \text{kg} + 1500 , \text{kg} = 5000 , \text{kg} \]
Now, using Newton's second law \(F = ma\) to find the force required:
\[ F = \text{Total Mass} \times \text{Acceleration} \]
\[ F = 5000 , \text{kg} \times 3 , \text{m/s}^2 = 15000 , \text{N} \]
Therefore, the force required to accelerate the train at \(3 , \text{m/s}^2\) is \(15000 , \text{N}\).
Next, if the last car (which has a mass of \(1500 , \text{kg}\)) suddenly detaches, we need to calculate the new total mass of the train:
\[ \text{New Total Mass} = \text{Mass of Lead Car} + \text{Mass of 2nd Car} \]
\[ \text{New Total Mass} = 2000 , \text{kg} + 1500 , \text{kg} = 3500 , \text{kg} \]
Since the force applied to the train remains \(15000 , \text{N}\), we can now calculate the new acceleration of the remaining train using \(F = ma\):
\[ a = \frac{F}{m} \]
\[ a = \frac{15000 , \text{N}}{3500 , \text{kg}} \approx 4.29 , \text{m/s}^2 \]
Therefore, after the last car detaches, the new acceleration of the remaining train will be approximately \(4.29 , \text{m/s}^2\).
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