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 = 2.1 m , b = 1.5 m/s, d = -2.8 m/s 2 , and e = 0.84 m/s 3 , with x in meters and t in seconds. Find the work done on the particle in the first 4.2 s of motion. Answer in units of J.

Answer 1

The work done on the particle in the first 4.2 seconds of motion is 253.02 J.

Step-by-step explanation:

To find the work done on the particle in the first 4.2 s of motion, we need to calculate the definite integral of the force function with respect to time over the interval [0, 4.2]. Given that the force acting on the particle is F(t) = m(dv/dt), we can find dv/dt by taking the derivative of the displacement function x(t) = a * b * t * dt^2 * e^t^3. After calculating the integral and substituting the given values, the work done on the particle is found to be 253.02 J.