Question

1. an object of mass 5 kg is moving on a smooth surface with a velocity of 6 m/s to the right. a constant force is applied for 3 s, giving the object a final velocity of 12 m/s to the left. what was the applied force?

Explanation:

Step1: Define Variables and Directions

Let right be positive. Initial velocity \( u = 6 \, \text{m/s} \), final velocity \( v = -12 \, \text{m/s} \) (negative as it's left), mass \( m = 5 \, \text{kg} \), time \( t = 3 \, \text{s} \).

Step2: Calculate Change in Momentum

Momentum change \( \Delta p = m(v - u) \).
Substitute values: \( \Delta p = 5(-12 - 6) = 5(-18) = -90 \, \text{kg·m/s} \).

Step3: Apply Impulse-Momentum Theorem

Impulse \( J = \Delta p \), and \( J = F \cdot t \). So \( F = \frac{\Delta p}{t} \).
Substitute: \( F = \frac{-90}{3} = -30 \, \text{N} \). The negative sign indicates force is to the left.

Answer:

The applied force is \( -30 \, \text{N} \) (or 30 N to the left).