Question

Que. A force F = -k/x^2 (x≠0) acts on a partical in a x-direction. Find the word done by this force in displacing the particle from x = +a to x = +2a. Here k is a positive constant.

Pls give correct answer-​

Answer 1

Answer :-

$\implies \boxed{\sf{ w = \frac{ - k}{2a \: } \: }} \: \bf{ N-m } \:$

To Find :-

→ Work done by given force .

Explanation :-

Given that

$\implies \: \sf{F = \frac{ - k}{ {x}^{2} } } \: \: \tt{N}$

it displacing +a to +2a

We know that

If force is variable then work done is -

$\implies \boxed{ \sf{w = \int_{initial}^{final} \: \: f \: dx \:} } \\ \\ \implies \sf{w = \int_{a}^{2a} \: \: \frac{ - k}{ {x}^{2} } \: dx \: } \\ \\ \implies \sf{ w = - k \: \int_{a}^{2a} \: {x}^{ - 2} \: dx \: } \\ \\ \: \: \: \: \: \: \: \: \: \: \: \because \boxed{\sf{ \int \: {x}^{n} dx = \frac{ {x}^{n + 1} }{n + 1} }} \\ \\ \implies \sf{ w = - k \: \bigg[ \frac{ {x}^{ - 2 + 1} }{ - 2 + 1} \bigg]_{a}^{2a} \: } \\ \\ \implies \sf{w = \cancel{ -} k \: \bigg[ \frac{ {x}^{ - 1} }{ \cancel{ - }1} \bigg]_{a}^{2a}} \: \\ \\ \implies \sf{ w = \:k \: \bigg[ \frac{ 1 }{ x} \bigg]_{a}^{2a}}$

taking limit

$\implies \sf{w = k \: \bigg[ \frac{ 1 }{ 2a} - \frac{1}{a} \bigg] \: } \\ \\ \implies \sf{w = k \: \frac{ - 1}{2a} } \\ \\ \implies \boxed{\sf{ w = \frac{ - k}{2a \: } \: }} \: \bf{ N-m }$

Answer 2

$\tt{\huge{\orange {------------- }}} S.D$

QUESTION :

Que. A force F = -k/x^2 (x≠0) acts on a partical in a x-direction. Find the word done by this force in displacing the particle from x = +a to x = +2a. Here k is a positive constant.

Que. A force F = -k/x^2 (x≠0) acts on a partical in a x-direction. Find the word done by this force in displacing the particle from x = +a to x = +2a. Here k is a positive constant.Pls give correct answer-

ANSWER :

Work done by this force is { - k } / { 2 a } N m.

CONCEPT USED :

Force acting on a Spring i.e, Spring Constant.

SOLUTION :

We have the following information given that F = [ - k / { x } ^ 2 ] Newton.

This force is displacing the particle, from X = + a to X = + 2 a.

We know that :

$Work \: Done = \int_{a}^{2a} \: \: \dfrac{-k }{ { x} ^ { 2 } }\: dx$

As :

$\int \: {x}^{n} dx = \frac{ {x}^{n + 1} }{n + 1}$

Using the above identity, we can state that :

$Work \: Done = - k \: \frac{ {x}^{ - 2 + 1} }{ - 2 + 1_{a}^{2a} }$

Now we need to cancel K

=>

$Work \: Done = k \frac{1}{2a} - \frac{1}{a}</p><p>$

This gives us our final answer :

Work done by this force is { - k } / { 2 a } N m..... [ A ]