Physics, asked by pappusath88, 1 year ago

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Answers

Answered by Anonymous
5

\Huge{\underline{\underline{\mathfrak{Question : }}}}

A force F = 20 + 10y acts on a particle in y direction where F is newton and y is metre. Find Work Done by this force to move the particle from y = 0 to y = 1m.

a) 30 J

b) 5 J

c) 25 J

d) 20 J

\Huge{\underline{\underline{\mathfrak{Answer \colon }}}}

\huge{\underline{\underline{\mathtt{Given \colon}}}}

  • Force acting on the object,F = 20 + 10y

  • Displacement from y = 0 to 1 m

To find

Work Done by the force in displacing the object

Since,

The force acting on the object is a variable force,we need to integrate the given function with respect to 'y'.

We Know that,

\displaystyle \boxed{\boxed{ \sf W = \int F.dy}}

On integrating F w.r.t to y,we get :

\boxed{\boxed{\sf W = 25 \ J}}

REFER TO THE ATTACHMENT

Thus,the work done by the force is 25 J___________(Option C)

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Answered by ShivamKashyap08
20

\huge{\bold{\underline{\underline{....Answer....}}}}

\huge{\bold{\underline{Given:-}}}

  • Force is given as F = 20 + 10y.
  • Displacement of particle from y = 0 to y = 1m.

\huge{\bold{\underline{Explanation:-}}}

\rule{300}{1.5}

As The Force is Changing w.r.t Displacement,

Therefore,

The applied Formula Should be,

\large{\boxed{\tt W = \displaystyle\int F.ds}}

\rule{300}{1.5}

\rule{300}{1.5}

As the Body is moving towards y - axis.

The Formula becomes,

\large{\tt \leadsto W = \displaystyle\int F.dy}

Substituting the values,

\large{\tt \leadsto W = \displaystyle\int (20 + 10y).dy}

Applying the limits,

\large{\tt  \leadsto W = \displaystyle\int^1_0 (20 + 10y).dy}

Now, Integrating,

\large{ \leadsto \displaystyle \tt  W = \left[ \dfrac{20y}{0+1} + \dfrac{10y^2}{1 + 1} \right]^1_0}

The above integration Formula used is :-

\large{\boxed{ \displaystyle\int \tt x^n dy = \dfrac{x^n + 1}{n + 1} \: \rightarrow n \neq- 1}}

Now, Simplifying the equation,

\large{ \leadsto  \displaystyle \tt  W = \left[ \dfrac{20y}{1} + \dfrac{10y^2}{2} \right]^1_0}

\large{ \leadsto  \displaystyle \tt W = \left[ \dfrac{20y}{1} + \dfrac{\cancel{10}y^2}{\cancel{2}} \right]^1_0}

\large{\tt \leadsto  W = \Bigg[ 20y + 5y^2 \Bigg]^1_0 }

Substituting the value of y values,

\large{\tt \leadsto  W = \Bigg[ 20 \times 1 + 5 \times (1)^2 - (20 \times 0 + 5 \times (0)^2)\Bigg] }

\large{\tt \leadsto  W = \bigg[ 20 \times 1 + 5 \times (1)^2 - 0\bigg] }

\large{\tt \leadsto  W = \bigg[ 20 \times 1 + 5 \times (1)^2 \bigg] }

\large{\tt \leadsto W = 20 + 5}

\huge{\boxed{\boxed{\tt W = 25 \: J}}}

So, the work done by the force is 25 Joules .

\rule{300}{1.5}

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