A 1630-kg car is traveling with a speed of 11.7 m/s. What is the magnitude of the horizontal net force that is required to bring the car to a halt in a distance of 54.2 m?

Question

BlackTom AI · Accepted Answer

To tackle this problem, I’ll walk through the physics step by step and then map the result to the provided options.
First, identify the known quantities: mass m = 1630 kg, initial speed u = 11.7 m/s, stopping distance s = 54.2 m, final speed v = 0 m/s.
Next, use the kinematic equation that relates speed, acceleration, and distance for a stopping (constant) acceleration: v^2 = u^2 + 2 a s. With v = 0, this becomes 0 = u^2 + 2 a s, so the acceleration a = -u^2 / (2 s).
Plugging in the numbers: u^2 = 11.7^2 = 136.89 (m^2/s^2), and 2s = 2 × 54.2 = 108.4 m. Therefore a = -136.89 / 108.4 ≈ -1.263 m/s^2. The negative sign indicates deceleration.
Now determine the net force via Newton’s second law: F = m a. Using the magnitude, F = 1630 kg × 1.263 m/s^2 ≈ 2060 N.
So the horizontal net force needed to stop the car in 54.2 m is about 2060 N. 
Comparing to the given answer choices: 1010 N, None of these, 1512 N, 2058 N, and 965 N, the value closest to 2060 N is 2058 N. This aligns with the calculation within rounding error.
If we consider the question’s provided “answer” field which lists 1512 N as the correct option, that would not match the physics-based result just derived; 1512 N would correspond to a smaller deceleration (a ≈ F/m ≈ 0.928 m/s^2) and would yield a stopping distance longer than 54.2 m (about 73.7 m). This highlights a possible discrepancy or miskey in the supplied answer field rather than in the physics itself.
In summary: the mathematically correct stopping force for the given data is approximately 2060 N, making 2058 N the best option among those listed, while 1512 N is not consistent with the stopping distance provided.

A 1630-kg car is traveling with a speed of 11.7 m/s. What is the magnitude of the horizontal net force that is required to bring the car to a halt in a distance of 54.2 m?单项选择题

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