Asked by Deadrea
The displacement of a particle with a mass of 225 g depends on time and is described by the function x(t) = -3.0t3 + 2.0t2 + 4.0t – 13. All constants are in SI units. Calculate the resultant force acting on the particle at time 2.3 s.
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Answered by
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To find the resultant force acting on the particle, we need to take the second derivative of the displacement function with respect to time:
x(t) = -3.0t3 + 2.0t2 + 4.0t – 13
v(t) = -9.0t2 + 4.0t + 4.0 (first derivative)
a(t) = -18.0t + 4.0 (second derivative)
Now we can evaluate the second derivative at t = 2.3 s:
a(2.3 s) = -18.0(2.3) + 4.0 = -36.4 m/s^2
The resultant force acting on the particle at time 2.3 s can be found using Newton's second law:
F = ma
F = (0.225 kg)(-36.4 m/s^2)
F = -8.19 N
The negative sign indicates that the force is acting in the opposite direction of the particle's motion. Therefore, the resultant force acting on the particle at time 2.3 s is 8.19 N, directed opposite to the particle's motion.
x(t) = -3.0t3 + 2.0t2 + 4.0t – 13
v(t) = -9.0t2 + 4.0t + 4.0 (first derivative)
a(t) = -18.0t + 4.0 (second derivative)
Now we can evaluate the second derivative at t = 2.3 s:
a(2.3 s) = -18.0(2.3) + 4.0 = -36.4 m/s^2
The resultant force acting on the particle at time 2.3 s can be found using Newton's second law:
F = ma
F = (0.225 kg)(-36.4 m/s^2)
F = -8.19 N
The negative sign indicates that the force is acting in the opposite direction of the particle's motion. Therefore, the resultant force acting on the particle at time 2.3 s is 8.19 N, directed opposite to the particle's motion.
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