A partile of mass 500g is moving in a circle of radius 1m with speed v (m/s) = 4t^2. The rate of work done on the particle at t = 2s is ?
Asked by dipanshusingla029
3rd April 2018,
5:46 PM
Answered by Expert
Answer:
Workdone dW = force × displacement = m(dv/dt)×dS
(dv/dt) is rate of change of velocity, that equals acceleration; dS is displacement
dW = m(dv/dt)×dS = m×(dS/dt)×dv = m×v×dv
we are given v = 4t2 ; hence dv = 8tdt
hence total workdone in 2 second is W = m ∫32t3dt (limits of integration 0 to 2)
hence W = 32m×(16/4) = 32 × 0.5× 4 = 64 J
Answered by Expert
3rd April 2018,
10:51 PM
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