Solution:
Given x = 3t -4t²+t³ differentiating v = dx/dt = 3-8t+ 3t²,
Initial speed = 3 m/s
Final speed = 3-32+48= 19 m/s
From Work Energy Theorem,
Work Done = Change in Kinetic Energy= ½×2×(19²-3²)= 352 J
A force act on a 2 kg particle such a way that position of the particle as a function of time is given by x = 3t -4t²+t³, Where x is in meter and t is in second. Work done during first 4 second is
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