Question

A force acts on a 2kg object so that it is position is given as a function of time as x=3t^2+5.what is the work done by this force in first 5 sec

Answer 1

Heyy here is ur answer...

Answer 2

Answer:

Explanation:

Work done against conservative force is give by=∆ke

There fore ∆KE IS = 1/2 M (Vf^2-Vi^2)

Wkt X= 3t^2 +5

Now DX/dt= 6t

Substitute 6t in ke

∆ke= 1/2 ×2(6×5^2- 6×0^2)

= 900j