Question

A single force acts on a particle-like object of mass m kg in such a way that the position of the object as a function of time is given by x = 3t - 4t2 + t3, with x in meters and t in seconds. Find the work done on the object by the force between t = 0 and time t. Express your answer in terms of the variables given.

Answer 1

Answer:

W = m * (-24t + 41t2 - 20t3 + 3t4)

Explanation:

The work is calculated multiplying the force by the change in distance.

The position of the object in t=0 is x=0, and the position in time t is x = 3t - 4t2 + t3, so the change in distance for time t is dx = 3t - 4t2 + t3

The force is calculated multiplying the mass of the object by its acceleration. The acceleration can be calculated derivating the expression of the distance two times (the first derivate gives us the velocity of the object):

v = 3 - 8t + 3t2

a = -8 + 6t

The acceleration in t=0 is -8, and the acceleration in time t is -8 + 6t, so to calculate the force, we need to use a mean acceleration between these two times (as it increases linearly): [(-8)+(-8+6t)]/2 = -8 + 3t

So the total force from time 0 to time t is F = m * (-8 + 3t)

Now we can calculate the work done by the force moving the object (W) multiplying the force F by the change in distance dx:

W = F * dx = m * (-8 + 3t) * (3t - 4t2 + t3) = m * (-24t + 41t2 - 20t3 + 3t4)

The work's unit is Joule, as all the other units are in SI: mass in kg, distance in meters and time in seconds.