Question 24: Work, Energy and Power & Circular Motion

Question

A body of mass 4 kg is placed on a plane at a point $P$ having coordinate $(3,4) \mathrm{m}$. Under the action of force $\overrightarrow{\mathrm{F}}=(2 \hat{i}+3 \hat{j}) \mathrm{N}$, it moves to a new point Q having coordinates $(6,10) \mathrm{m}$ in 4 sec . The average power and instanteous power at the end of 4 sec are in the ratio of :

Accepted Answer

$$\begin{aligned}
& \langle\mathrm{p}angle=\frac{(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}) \cdot(3 \hat{\mathrm{i}}+6 \hat{\mathrm{j}})}{4}=6 \
& \overrightarrow{\mathrm{a}}=\left(\frac{\overrightarrow{\mathrm{F}}}{\mathrm{~m}}=\frac{1}{2} \hat{\mathrm{i}}+\frac{3}{4} \hat{\mathrm{j}}ight) \
& \overrightarrow{\mathrm{v}} 	ext { at } \mathrm{t}=4 \mathrm{sec}=\left(\frac{1}{2} \hat{\mathrm{i}}+\frac{3}{4} \hat{\mathrm{j}}ight) 	imes 4=(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}) \
& P_{	ext {ins }}=(2 \hat{\mathrm{i}}+3)(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}})=13 \
& \frac{\langle\mathrm{P}angle}{P_{	ext {ins }}}=\frac{6}{13}
\end{aligned}$$
Note: Given data is not matching.
$$\begin{aligned}
& S=u t+\frac{1}{2} a t^2 \
& S=0+\frac{1}{2} \frac{(2 \hat{i}+3 \hat{j})}{4}(4)^2=4 \hat{i}+6 \hat{\mathrm{j}}
\end{aligned}$$
If $\overrightarrow{\mathrm{r}}_{\mathrm{i}}=3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}$ then $\overrightarrow{\mathrm{r}}_{\mathrm{f}}=7 \hat{\mathrm{i}}+10 \hat{\mathrm{j}}$
But Final position given in the question is $(6,10)$.

Answer

4 : 3

Answer

13 : 6

Answer

1 : 2

Answer

6 : 13

Work, Energy and Power & Circular Motion

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