Question

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★ Qᥙเᤁᤁ :

➽ A force F = ( 2 + x ) acts on a particle in x-direction where F is in newton and x in meter.Find the work done by this force during a displacement from x = 1.0 m to x = 2.0 m

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Answer 1

Work done = F.x

F.x= (2+x)x = 2x+x²

at x= 1 , W= 2(1)+1² = 3J

at x= 2, W = 2(2)+2² = 8 J

Net work done = 8-3 = 5J

You can ask me questions directly

Answer 2

${ \maltese \: \: \: { \underline{ \purple{ \underline{ \pmb{ \bf{Given \: that : }}}}}}}$

✰ A force F = ( 2 + x ) acts on a particle in x-direction where F is in newton and x in m

$\: { \maltese \: \: \: { \underline{ \purple{ \underline{ \pmb{ \bf{To \: Find : }}}}}}}$

✰ Find the work done by this force during a displacement from x = 1.0 m to x = 2.0 m

$\: { \maltese \: \: \: { \underline{ \purple{ \underline{ \pmb{ \bf{ Required \: Solution : }}}}}}}$

➽ The work done by the force during the given displacement is 3.5 joules

$\: { \maltese \: \: \: { \underline{ \purple{ \underline{ \pmb{ \bf{Full \: Solution : }}}}}}}$

➤ Here in this question we have been stated that a force F which is equal to 2 + x acts on a particle in the x - direction where F is in newton and x in meter.

✰ Now, we have been asked that we should find the work done during the displacement from x = 1.0 m to x = 2.0 m

★ Now We know that :

The force is variable and that we shall find the work in small displacement x to x + dx and then integrate it to find the work done the by force is dW = F dx = ( x + 2 ) dx

So, the work done here in the given displacement id dw

★ The work done due to a variable force is given by :

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: { \footnotesize{\bullet \: { \underline{ \boxed{ \bf{ W = \int \limits ^{r_{2}}_{r_1}dW = \int \limits ^{r_{2}}_{r_1} F . \: dx \: \: }}}}}}$

Here :

• W Stands dor Work done
• R₁ is 1.0 m
• R₂ is 2.0 m
• F is 2 + x

Integrating We Get :

$: \implies{\bf{ W = \int \limits ^{2.0}_{1.0}dW = \int \limits ^{2.0}_{1.0} F . \: dx }} \\ \\ \\ : \implies{\bf{ W = \int \limits ^{2.0}_{1.0}dW = \int \limits ^{2.0}_{1.0} F . \: dx }} \\ \\ \\ : \implies{\bf{ W = \bigg[ \frac{ {x}^{2} }{2} + 2x \bigg ] {}^{ {}^{ {}^{ {}^{ {}^{2} } } } } }} \\ \\ \\ : \implies{\bf{W = \bigg[ \frac{{2}^{2} }{2} + 2(2) \bigg] }} \\ \\ \\ : \implies{\bf{ W = \frac{ \cancel4}{ \cancel2} + 4 - \bigg[ \frac{1}{2} + 2 \bigg]}} \\ \\ \\ : \implies{\bf{ W = 2 + 4 - \frac{1}{2} - 2}} \\ \\ \\ : \implies{\bf{ W =6 - 2 - \frac{1}{2} }} \\ \\ \\ : \implies{ \purple{ \underline{ \boxed{ \pmb{\bf{ W = 3.5 \: Joules}}}}}}$

$\: { \maltese \: \: \: { \underline{ \purple{ \underline{ \pmb{ \bf{Therefore : }}}}}}}$

➽ The work done by the force in the given displacement is 3.5 Joules

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