A single force acts on a 2.0 kg particle-like object in such a way that the object's position as a function of time is given by x = 3.0t - 4.0t^2 + 1.0t^3, where x is in meters and t is in seconds. Calculate the work done by the force on the object from t = 0 to t = 5.0 s.

 To calculate the work done on the object by the force, we first need to find the force acting on the object as a function of time. We can do this by using the given position function \( x(t) \) and employing Newton's second law:

\( F = m \cdot a \)

where \( F \) is the force, \( m \) is the mass, and \( a \) is the acceleration of the object. We know that the acceleration is the second derivative of the position with respect to time. Let's find the acceleration first.

The position function is given by:

\( x(t) = 3.0t - 4.0t^2 + 1.0t^3 \)

The first derivative with respect to time \( t \) gives the velocity:

\( v(t) = x'(t) = \frac{dx}{dt} = 3.0 - 8.0t + 3.0t^2 \)

The second derivative gives the acceleration:

\( a(t) = v'(t) = \frac{dv}{dt} = \frac{d^2x}{dt^2} = -8.0 + 6.0t \)

Now we multiply the acceleration by the mass to find the force as a function of time:

\( F(t) = m \cdot a(t) = 2.0 \cdot (-8.0 + 6.0t) = -16.0 + 12.0t \)

The work done by a variable force along a path is given by the integral of the force along that path. Since the object moves in the x-direction, we can write the work \( W \) as follows:

\( W = \int F \cdot dx \)

But since we have \( F \) as a function of \( t \), and we know that \( dx = v dt \), we need to express work done in terms of time:

\( W = \int_{t_i}^{t_f} F(t) \cdot v(t) dt \)

Where \( t_i \) is the initial time and \( t_f \) is the final time.

Substitute the expressions we found for \( F(t) \) and \( v(t) \):

\( W = \int_{0}^{5} (-16.0 + 12.0t) \cdot (3.0 - 8.0t + 3.0t^2) dt \)

Now, we need to perform the integration:

\( W = \int_{0}^{5} (-16 \cdot 3 + 12 \cdot 3t + 16 \cdot 8t - 12 \cdot 8t^2 - 16 \cdot 3t^2 + 12 \cdot 3t^3) dt \)

\( W = \int_{0}^{5} (-48 + 36t + 128t - 96t^2 - 48t^2 + 36t^3) dt \)

\( W = \int_{0}^{5} (-48 + 164t - 144t^2 + 36t^3) dt \)

Integrating term by term within the limits from 0 to 5 gives:

\( W = [(-48t + 82t^2 - \frac{144}{3}t^3 + \frac{36}{4}t^4)]_{0}^{5} \)

\( W = [(-48t + 82t^2 - 48t^3 + 9t^4)]_{0}^{5} \)

\( W = (-48 \cdot 5 + 82 \cdot 25 - 48 \cdot 125 + 9 \cdot 625) - 0 \)

\( W = (-240 + 2050 - 6000 + 5625) \)

\( W = 1435 \) joules.

Therefore, the work done by the force on the object from \( t = 0 \) to \( t = 5.0 \) seconds is 1435 joules.