A single force acts on a 1.1 kg particle-like object in such a way that the position of the object as a function of time is given by x = 4.1t - 3.6t2 + 4.5t3, with x in meters and t in seconds. Find the work done on the object by the force from t = 0 to t = 8.6 s.
work = integral f dx = change in kinetic energy if this is the only force
v =dx/dt = 4.1 -7.2 t + 13.5 t^2
at t = 0, v = 0
at t = 8.6
v = 4.1 - 7.2(8.6) + 13.5 (8.6)^2
= 941 m/s^2
Ke = .5 m v^2 = .5(1.1)(942)^2
= 486,642 Joules
To find the work done on the object by the force from t = 0 to t = 8.6 s, we need to calculate the area under the curve of the force-position graph.
First, we need to find the force acting on the object. The force on an object is given by the derivative of its position with respect to time. So we differentiate the position function x(t) = 4.1t - 3.6t^2 + 4.5t^3 with respect to time to find the force function F(t).
x(t) = 4.1t - 3.6t^2 + 4.5t^3
Now differentiate with respect to t:
v(t) = dx/dt = 4.1 - 7.2t + 13.5t^2
Next, differentiate the velocity function v(t) to find the acceleration function a(t).
a(t) = dv/dt = -7.2 + 27t
Finally, the force function F(t) is given by multiplying the acceleration function a(t) by the mass of the object (1.1 kg):
F(t) = m * a(t) = 1.1 * (-7.2 + 27t)
Now that we have the force function F(t), we need to calculate the work done by integrating the force over the interval from t = 0 to t = 8.6s.
W = ∫[0, 8.6] F(t) dt
= ∫[0, 8.6] (1.1 * (-7.2 + 27t)) dt
= 1.1 * ∫[0, 8.6] (-7.2 + 27t) dt
= 1.1 * [-7.2t + 13.5t^2/2] evaluated from t = 0 to t = 8.6
= 1.1 * ([-7.2 * 8.6 + 13.5 * (8.6)^2/2] - [-7.2 * 0 + 13.5 * (0)^2/2])
Now, plug in the values and calculate the expression:
W = 1.1 * ([-7.2 * 8.6 + 13.5 * (8.6)^2/2] - [0])
Simplifying, we get:
W = 1.1 * (-61.92 + 639.66)
W = 1.1 * 577.74
W ≈ 635.51 Joules
Therefore, the work done on the object by the force from t = 0 to t = 8.6 s is approximately 635.51 Joules.