Net Force on a Particle in Motion

The Scenario

• Question: Determine the net force acting on a 0.150 kg particle moving along the x-axis at t = 3.40 seconds. The particle's position function is x(t) = -13.00 + 2.00t + 4.00t² - 3.00t³, with x in meters and t in seconds.

Solution Explanation

Step 1: Finding the Velocity

• Position Function: x(t) = -13.00 + 2.00t + 4.00t² - 3.00t³
• Velocity Calculation: v(t) = dx/dt = 2.00 + 8.00t - 9.00t²

Step 2: Calculating the Acceleration

• Acceleration Calculation: a(t) = d²x/dt² = 8.00 - 18.00t
• Acceleration at t = 3.40 seconds: a(3.40) = 8.00 - 18.00 * 3.40
• In Unit-Vector Notation: a(3.40) = (8.00 - 18.00 * 3.40)i m/s²

Step 3: Determining the Net Force

• Mass of Particle: 0.150 kg
• Force Calculation: F = m * a = 0.150 kg * (8.00 - 18.00 * 3.40)i
• Net Force at t = 3.40 seconds: F ≈ -7.98i N

Conclusion

• At 3.40 seconds, the particle experiences a net force of approximately -7.98 Newtons in the negative x-direction.
• This force calculation is crucial for understanding the dynamics of the particle's motion along the x-axis.