Question

if a car with a mass of 1,200 kg traveling westward at 30 m/s is slowed to a stop in 3 s, then what was the net force acting on the car? take east to be positive and west to be negative. a - 12,000 n b 36,000 n c 3600 n d 12,000 n

Explanation:

Step1: Calculate the acceleration

Use the formula \(a=\frac{v - u}{t}\), where \(u=- 30\ m/s\) (initial velocity, westward is negative), \(v = 0\ m/s\) (final velocity), \(t = 3\ s\).
\(a=\frac{0-(-30)}{3}=\frac{30}{3}=10\ m/s^{2}\) (the negative sign for force will come from the direction consideration in force formula)

Step2: Calculate the net force

Use Newton's second law \(F = ma\), where \(m = 1200\ kg\), \(a=\frac{v - u}{t}\). Substituting the values \(F=1200\times\frac{0 - (- 30)}{3}\)

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Since the car is decelerating (slowing down) while moving westward (negative direction as per convention), the force is opposite to the direction of motion. So \(F=-12000\ N\)

Answer:

A. - 12,000 N