NEET Physics Work, Energy, Power & Collision / कार्य, ऊर्जा और शक्ति NEET PYQ-Work Energy Power and Collision

A force acts on a 3.0 g particle in such a way that the position of the particle as a function of time is given by \[x=3t-4{{t}^{2}}+{{t}^{3}},\] where \[x\] is in metre and \[t\] in second. The work done during the first 4 s is: [AIPMT 1998]

A) 570 mJ

B) 450 mJ

C) 490 mJ

D) 528 mJ

Correct Answer: D

Solution :

Key Idea: Work done during the first 4s is equal to gain in kinetic energy.

We have given,

\[x=3t-4{{t}^{2}}+{{t}^{3}}\]

So, velocity

\[v=\frac{dx}{dt}=3-8\,t+3\,{{t}^{2}}\]

At \[t=0,\]\[{{v}_{1}}=3-0+0=3\,m/s\]

At \[t=4\,s,\] \[{{v}_{2}}=3-8\times 4+3\times {{4}^{2}}\]

\[=3-32+48=19\,m/s\]

Now work done during \[t=0\] and \[t=4s\]= gain in kinetic energy

\[=\frac{1}{2}mv_{2}^{2}-\frac{1}{2}mv_{1}^{2}=\frac{1}{2}m(v_{2}^{2}-v_{1}^{2})\]

\[=\frac{1}{2}\times 3\times {{10}^{-3}}\,[{{(19)}^{2}}-{{(3)}^{2}}]\]

\[[\text{Using}\,\,{{a}^{2}}-{{b}^{2}}=(a+b)\,(a-b)]\]

\[=1.5\times {{10}^{-3}}\times [\,(19+3)\,(19-3)]\]

\[=1.5\times {{10}^{-3}}\times 22\times 16\]

\[=528\times {{10}^{-3}}\,J\]

\[=528\,mJ\]