Question

A time-varying net force acting on a 2.6 kg particle causes the object to have a displacement given by x = a + b t + d t2 + e t3 , where a = 1.1 m , b = 1.8 m/s, d = −3.1 m/s 2 , and e = 1.1 m/s 3 , with x in meters and t in seconds. Find the work done on the particle in the first 4.4 s of motion. Answer in units of J.

Answer 1

Answer:

The answer is: 1913.5158 J

Explanation:

First we need to derivate the displacement equation:

$x = 1.1+1.8t-3.1t^{2} +1.1t^{3} \\\frac{dx}{dt} = 1.8-6.2t+3.3t^{2} \\\frac{d^{2} x}{dt^{2} } = -6.2+6.6t$

Second, we need to know the 2nd Law of Newton:

$F = m*a = m\frac{d^{2} x}{dt^{2} }$

So, we have:

$F = 2.6*(-6.2+6.6t)$

Third, to calculate the work done on the particle we need the next equation:

$W= \int\limits^a_b {F} \, dx$

Then, we have everything to replace in the variables(dx is replaced by the first derivative) :

$W= \int\limits^{4.4}_{0} {2.6*(-6.2+6.6t)*(1.8-6.2t+3.3t^{2})} \, dt$

And finally we have as a result:

W = 1913.5158 J