Question

Under the action of a force a 2Kg body moves such that it's position 'x' in meters as a function of time 't' in seconds given by ; x = t2 /2 . the work done by the force in the first 5 seconds is


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Answer 1

Answer:

Under the action of a force, a 2 kg body moves - such that its position x as a function of time t is. ... The work done by the force in first two. seconds is. (1) 1600 J.

Answer 2

Given: Under the action of a force a 2 kg body moves such that it's position 'x' in meters as a function of time 't' in seconds given by ; x = t² /2 .

To find: The work done by the force in the first 5 seconds.

Solution:

Method 1

                                In order to obtain the values of work done, we first differentiate x till we obtain acceleration for the value of force. So,

                       dx/ dt = t² / 2

                              v = 2t / 2

                       dv/ dt = 2t/ 2

                              a = 2/2

                              a = 1

We know that for work done:

                      W = F.s

                      W = ma × (t²/ 2)⁵₀

                      W = 2 × 1 × 25/2

                      W = 25 J

Method 2

                                By the work energy theorem, we know that:

                      W = 1/2 × m (Vf² - Vi²)

For velocity, we differentiate position 'x':

                   dx/ dt = t²/ 2

                          v = 2t/2

                          v = t

We find the work done as:

                    W = 1/2 × 2 × (Vf² - Vi²)

                    W = 5² - 0²

                    W = 25 J.

Hence, the work done by the force in the first 5 seconds is 25 J.