Question

A particle of mass 1 kg moving with variable vs=kt find mean kinetic energy from t=1 to t=2 sec

Answer 1

The mean kinetic energy of the particle from t=1s to t=2s is equal to 7k²/6 J.

• v = kt , m = 1 kg and KE = mv²/2
• KE = k²t²/2
• The mean kinetic energy is equal to ∫KE.dt / ∫dt

• Therefore the mean kinetic energy = ∫0.5k²t².dt / ∫dt

                                                                   = 0.5k².((t2³-t1³)/3) / (t2-t1)

• In the above expression t1 and t2 are the limits of integration. For the above question, t1=1s and t2=2s
• Hence, the mean kinetic energy equals 0.5k²((2³-1³)/3) / (2-1) = 7k²/6 J