Question

A 0.150 kg particle moves along an x axis according to x(t) = -13.00 + 2.00t + 4.00t² - 3.00t³, with x in meters and t in seconds. In unit-vector notation, what is the net force acting on the particle at t = 3.40 s?

Answer 1

The net force acting on the particle at t = 3.40 s is -7.98 N in the negative x-direction, found by deriving the acceleration from the position function and applying Newton's second law.

Step-by-step explanation:

To determine the net force acting on a particle of mass 0.150 kg moving along an x-axis according to the position function x(t) = -13.00 + 2.00t + 4.00t² - 3.00t³, we first need to find the acceleration of the particle at the specific time t = 3.40 s.

The acceleration is the second derivative of the position function concerning time. Calculating this, we get:

a(t) = d²x/dt² = 8.00 - 18.00t

Plugging in t = 3.40 s:

a(3.40 s) = 8.00 - 18.00(3.40) = 8.00 - 61.20 = -53.20 m/s²

Now, the net force can be found using Newton's second law, F = ma.

F(3.40 s) = (0.150 kg)(-53.20 m/s²) = -7.98 N

Thus, the force in unit-vector notation is F = -7.98 N Ħ where Ħ represents the unit vector along the x-axis.