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Grade 12th passAIPMT

A particle of mass 3kg is moving along x-axis and its position at time t is given by eqn x=(2t

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Profile image of Srivarsan
8 Years agoGrade 12th pass
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1 Answer

Profile image of Devendra Dansena
8 Years ago
By the given eqn x=2t^2+5 We can get the acceleration by double differentiating the eqn d^2x/dt^2 = acceleration= 4m/s^2 Now by 2nd low of equation S=ut+1/2at^2 Let initial velocity be = 0 Then we getS=1/2*4*9= 18m ( t=3)Now work done = FS =m*a*sW= 3*4*18= 216 joules

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