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 = 2 m , b = 1.4 m/s, d = −1.8 m/s 2 , and e = 1.2 m/s 3 , with x in meters and t in seconds. Find the work done on the particle in the first 3.3 s of motion. Answer in units of J.

W = 1579.94J

Explanation:

x = a + bt + dt² + et³

First, we find acceleration:

v = dx/dt = b + 2dt + 3et²

a = dv/dt = 2d + 6et

d = -1.8 m/s², e = 1.2 m/s³

a = 2*(-1.8) + 6*1.2*t

a = -3.6 + 7.2t

Force, F is given as:

F = m * a

F = 2.6 * (-3.6 + 7.2t)

F = -9.36 + 18.72t

When t = 3.3 secs:

F = -9.36 + (18.72*3.3)

F = 52.42N

x = a + bt + dt² + et³

Inputting values of a, b, d, e:

x = 2 + (1.4*3.3) + (-1.8*3.3²) + (1.2*3.3³)

x = 2 + 4.62 - 19.602 + 43.1244

x = 30.1424m

Therefore, Work done, W is

W = F * x

W = 52.42 * 30.1424

W = 1579.94J