JEE Main 2019 Physics question
A force acts on a 2 kg object so that its position is given as a function of time as . What is the work done by this force in first 5 seconds?
Answer: (B)
Solution
v = 6t = 30 m/s at t = 5 s; W = ΔKE = ½(2)(900) = 900 J.