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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Work, Power and Energy
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Displacement of a body of mass 2 kg varies with time as $s=t^2+2t$.Where s in meters t in seconds. The work done by the force acting an body during the time t=2s to t=4s is

\[(a)\;36\;J \quad (b)\;64\;J \quad (c)\;100\;J \quad (d)\;120\;J\]

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Answer: 64 J
The Work-Energy theorem for a variable force states that $K_f - K_i = W$, where $K_f$ and$ K_i$ are the final and initial Kinteic energies, with the corresponding velocities $v_f$ and $v_i$, and initial and final times of $t_i = 2s$ and $t_f = 4s$.
Given $s = t^2 + 2t \rightarrow v = \large\frac{ds}{dt}$$ = \large\frac{d (t^2+2t)}{dt}$$ = 2t + 2$
$\Rightarrow K_f = \large\frac{1}{2}$ $m v_f^2 $$= \large\frac{1}{2} $$ \times 2 (2 \times 4 + 2)^2$$ = 100J$
$\Rightarrow K_i = \large\frac{1}{2}$ $m v_i^2 $$= \large\frac{1}{2} $$ \times 2 (2 \times 2 + 2)^2$$ = 36J$
$\Rightarrow W = K_f - K_i = 100 - 36 = 64J$
answered Jul 17, 2013 by meena.p
edited Aug 25, 2014 by balaji.thirumalai
 

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