A 0.150 kg particle moves along an x axis according to x(t) = -13.00 + 2.00t + 4.50t2 - 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 = 3.60 s?
To find the net force acting on the particle at t = 3.60 s, we need to calculate the second derivative of x with respect to time (t), and then multiply it by the mass of the particle.
Given that x(t) = -13.00 + 2.00t + 4.50t^2 - 3.00t^3, we can find the first derivative, v(t), by differentiating x(t) with respect to t:
v(t) = d/dt(x(t))
= d/dt(-13.00 + 2.00t + 4.50t^2 - 3.00t^3)
= 2.00 + 9.00t - 9.00t^2
Next, we find the second derivative, a(t), by differentiating v(t) with respect to t:
a(t) = d/dt(v(t))
= d/dt(2.00 + 9.00t - 9.00t^2)
= 9.00 - 18.00t
Now that we have the acceleration, we can find the net force by multiplying the acceleration by the mass of the particle:
m = 0.150 kg (mass of the particle)
a(t) = 9.00 - 18.00t (acceleration at t = 3.60 s)
F_net = m * a(t = 3.60 s)
= 0.150 kg * (9.00 - 18.00 * 3.60 )
= 0.150 kg * (9.00 - 64.80)
= 0.150 kg * (-55.80)
= -8.37 N
Therefore, the net force acting on the particle at t = 3.60 s is -8.37 N in the direction opposite to the positive x-axis.
Compute the second derivative x"(t) at t = 3.60 s. That will be the acceleration at that time.
x'(t) = 2 + 9 t - 9 t^2
x"(t) = 9 - 18 t
x"(t=3.6) = -55.8 m/s^2
Multiply that by the mass for the force. It will be in the -x direction.