Question

When a force (8hati+4hatj) newton displaces a particle through (3hati-3hatj) metre, the power is 0.6W. The time of action of the force is

Answer 1

$\underline{\boxed{ \huge\purple{ \rm{Answer}}}} \\ \\ \star \: \rm \vec{F} = 8 \hat{i} + 4 \hat{j} \\ \\ \star \rm \: \vec{d} = 3 \hat{i} - 3 \hat{j} \\ \\ \star \rm \: w = \vec{F} \: { \tiny{ \bullet}} \: \vec{d} = 24 - 12 = 12 \: J \\ \\ \implies \rm \: power = \frac{work}{time} \\ \\ \therefore \rm \: time = \frac{w}{p} = \frac{12}{0.6} = \red{20 \: s}$

Answer 2

$\huge \fcolorbox{red}{pink}{Solution :)}$

Given ,

• Force = $\sf 8 \hat{i}+ 4 \hat{j}$

• Displacement = $\sf 3 \hat{i} - 3 \hat{j}$

• Power = $\sf 0.6 \: watt$

We know that ,

The dot product of force and displacement is called work

$\large \mathtt{ \fbox{Work = \vec{F} . \vec{D}}}$

Substitute the known values , we get

$\sf \hookrightarrow Work = (8 \hat{i}+ 4 \hat{j}) \: . \: (3 \hat{i} - 3 \hat{j}) \\ \\ \sf \hookrightarrow Work = (8 \times 3) + (4 \times ( - 3)) \\ \\ \sf \hookrightarrow Work = 24 - 12 \\ \\ \sf \hookrightarrow Work =12 \: \: joule$

We know that ,

The rate of work done is called power

$\large \mathtt{ \fbox{Power = \frac{Work \: done}{Time}} }$

Substitute the known values , we get

$\sf \hookrightarrow Time = \frac{12}{0.6} \\ \\ \sf \hookrightarrow Time = 20 \: \: sec$

Hence , the time of action of the force is 20 sec

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