Physics · Work done by a constant force and a variable force, kinetic and potential energies, work-energy theorem, power
A force acts on a particle in such a way that position of the particle as a func
A force acts on a \( 3 g \) particle in such a way that position of the particle as a function of time is given by \( \boldsymbol{x}=\mathbf{3} \boldsymbol{t}- \) \( 4 t^{2}+t^{3}, \) where \( x \) is in metre and \( t \) is in sec. The work done during the first \( 4 s \) is
- A. 570 mJ
- B. 450 mJ
- C. \( 490 \mathrm{mJ} \)
- D. 528 mJ
Step-by-step solution
The work done is equal to the change in kinetic energy. Velocity v = dx/dt = 3 - 8t + 3t^2. At t=0, v0=3 m/s; at t=4 s, v=19 m/s. Mass m=3 g=0.003 kg. ΔKE = 1/2 * 0.003 * (19^2 - 3^2) = 0.0015 * (361 - 9) = 0.0015 * 352 = 0.528 J = 528 mJ.
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