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Grade 12th passMechanics

a force act on a 2.80 kg particle in such a way that the position of the particle as a funcation of time is given by x=3t-4t^2+t^3, where x is in meter and t is in second. (a) find the work done by the force during the first 4.0 s (b) at what instsntineous rate is the force doing work on the particle at the instant t=3 s?

Profile image of Faizan ali
8 Years agoGrade 12th pass
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2 Answers

Profile image of Arun
8 Years ago
The speed of the particle is : v = dx/dt = 3.0 - 8t + 3 t^2. 
Then the kinetic energy at t=0 is: Ek(0) = 0.5*2.8*9 = 12.6 J ; 
the kinetic energy at t = 4.0 s is : Ek(2) =0.5*2.8*361 = 505.4 J. 
The work done by the force is: 
W = Ek(2) - Ek(0) = 505.4 + 12.6 = 518 J
Profile image of Khimraj
8 Years ago
Apply work energy theorem.
velocity v = dx/dt = 3 – 8t +3t2
initial KE = ½*2.8*32 = 12.6
final KE = ½*2.8*192 = 505.4
work done = change in KE = 505.4 – 12.6 = 492.8J