A 3.17×103 kg train is stopped at a station. The train must go around a 113 m diameter corner right after it leaves the station. As the train leaves the station it provides a constant forward force of 1.79×103 N. What is the train's acceleration after 11.0 s?
Question
A 3.17×103 kg train is stopped at a station. The train must go around a 113 m diameter corner right after it leaves the station. As the train leaves the station it provides a constant forward force of 1.79×103 N. What is the train's acceleration after 11.0 s?
Solution
To solve this problem, we need to use Newton's second law of motion, which states that the acceleration of an object is equal to the net force acting on it divided by its mass.
Step 1: Identify the given values.
- Mass (m) = 3.17×10^3 kg
- Force (F) = 1.79×10^3 N
- Time (t) = 11.0 s
Step 2: Use Newton's second law to calculate acceleration.
- Acceleration (a) = Force / Mass
- a = F / m
- a = 1.79×10^3 N / 3.17×10^3 kg
- a = 0.564 m/s^2
So, the train's acceleration after 11.0 s is 0.564 m/s^2.
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