Work Power And Energy Ques 5
5 A particle which is experiencing a force, is given by $\mathbf{F}=3 \hat{\mathbf{i}}-12 \hat{\mathbf{j}}$, undergoes a displacement of $\mathbf{d}=4 \hat{\mathbf{i}}$. If the particle had a kinetic energy of $3 \mathrm{~J}$ at the beginning of the displacement, what is its kinetic energy at the end of the displacement?
(2019 Main, 10 Jan II)
(a) $9 \mathrm{~J}$
(b) $15 \mathrm{~J}$
(c) $12 \mathrm{~J}$
(d) $10 \mathrm{~J}$
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Answer:
Correct Answer: 5.( b )
Solution:
Formula:
- We know that, work done in displacing a particle at displacement $\mathbf{d}$ under force $\mathbf{F}$ is given by
$ \Delta W=\mathbf{F} \cdot \mathbf{d} $
By substituting given values, we get
$ \begin{array}{lc} \Rightarrow & \Delta W=(3 \hat{\mathbf{i}}-12 \hat{\mathbf{j}}) \cdot(4 \hat{\mathbf{i}}) \\ \Rightarrow & \Delta W=12 \mathrm{~J} \end{array} $
Now, using work-energy theorem, we get work done $(\Delta W)=$ change in kinetic energy $(\Delta K)$
or
$ \Delta W=K_2-K_1 $
Comparing Eqs. (i) and (ii), we get
$ \begin{aligned} K_2-K_1 & =12 \mathrm{~J} \\ K_2 & =K_1+12 \mathrm{~J} \end{aligned} $
or Given, initial kinetic energy, $K_1=3 \mathrm{~J}$
$\therefore$ Final kinetic energy, $K_2=3 \mathrm{~J}+12 \mathrm{~J}=15 \mathrm{~J}$
BETA
